1 integer is 7 times another. If the product of the 2 integers is 448, then find the integers. Let the first integer be x and the second integer be y. We have the following two equations: [LIST=1] [*]x = 7y [*]xy = 448 [/LIST] Substitute (1) into (2), we have: (7y)y = 448 7y^2 = 448 Divide each side by 7 y^2 = 64 y = -8, 8 We use 8, since 8*7 = 56, and 56*8 =448. So the answer is [B](x, y) = (8, 56)[/B]

10 times the first of 2 consecutive even integers is 8 times the second. Find the integers. Let the first integer be x. Let the second integer be y. We're given: [LIST=1] [*]10x = 8y [*]We also know a consecutive even integer means we add 2 to x to get y. y = x + 2 [/LIST] Substitute (1) into (2): 10x = 8(x + 2) Multiply through: 10x = 8x + 16 To solve for x, [URL='https://www.mathcelebrity.com/1unk.php?num=10x%3D8x%2B16&pl=Solve']we type this equation into our search engine[/URL] and we get: [B]x = 8[/B] Since y = x + 2, we plug in x = 8 to get: y = 8 + 2 [B]y = 10 [/B] Now, let's check our work. Does x = 8 and y = 10 make equation 1 hold? 10(8) ? 8(10) 80 = 80 <-- Yes!

149 cars are waiting to take a ferry across the channel each ferry can only hold 18 cars how many trips will it take to get all the cars across Number of trips = Total Cars / Cars Per ferry trip Number of trips = 149/18 Number of trips = 8.28 trips We round up to the next integer and we have [B]9 trips[/B]

Let x be the first even integer. That means the next consecutive even integer must be x + 2. Set up our equation: x + (x + 2) = 118 Group x terms 2x + 2 = 118 Subtract 2 from each side 2x = 116 Divide each side by 2 x = 58 Which means the next consecutive even integer is 58 + 2 = 60 So our two consecutive even integers are [B]58, 60[/B] Check our work: 58 + 60 = 118

2 consecutive odd integers such that their product is 15 more than 3 times their sum. Let the first integer be n. The next odd, consecutive integer is n + 2. We are given the product is 15 more than 3 times their sum: n(n + 2) = 3(n + n + 2) + 15 Simplify each side: n^2 + 2n = 6n + 6 + 15 n^2 + 2n = 6n + 21 Subtract 6n from each side: n^2 - 4n - 21 = 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2-4n-21%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']Type this problem into our search engine[/URL], and we get: n = (-3, 7) If we use -3, then the next consecutive odd integer is -3 + 2 = -1. So we have [B](-3, -1)[/B] If we use 7, then the next consecutive odd integer is 7 + 2 = 9. So we have [B](7, 9)[/B]

237 what is the place value of 3 Place value for integers with no decimals from right to left is: 7 is the ones digit 3 is the [B]tens digit[/B]

2 consecutive even integers such that the smaller added to 5 times the larger gives a sum of 70. Let the first, smaller integer be x. And the second larger integer be y. Since they are both even, we have: [LIST=1] [*]x = y - 2 <-- Since they're consecutive even integers [*]x + 5y = 70 <-- Smaller added to 5 times the larger gives a sum of 70 [/LIST] Substitute (1) into (2): (y - 2) + 5y = 70 Group like terms: (1 + 5)y - 2 = 70 6y - 2 = 70 [URL='https://www.mathcelebrity.com/1unk.php?num=6y-2%3D70&pl=Solve']Typing 6y - 2 = 70 into our search engine[/URL], we get: [B]y = 12 <-- Larger integer[/B] Plugging this into Equation (1) we get: x = 12 - 2 [B]x = 10 <-- Smaller Integer[/B] So (x, y) = (10, 12)

3 consecutive odd integers such that thrice the middle is 15 more than the sum of the other 2. [LIST] [*]Let the first integer be n [*]The next odd one (middle) is n + 2. [*]The next odd one is n + 4 [/LIST] We are given 3(n + 2) = n + n + 4 + 15. Simplifying, we get: 3n + 6 = 2n + 19 [URL='http://www.mathcelebrity.com/1unk.php?num=3n%2B6%3D2n%2B19&pl=Solve']Plugging that problem[/URL] into our search engine, we get n = 13. So the next odd integer is 13 + 2 = 15 The next odd integer is 15 + 2 = 17

4 consecutive integers such that the sum of the first 3 integers is equal to the 4th Let n be our first consecutive integer. [LIST=1] [*]n [*]n + 1 [*]n + 2 [*]n + 3 [/LIST] The sum of the first 3 integers is equal to the 4th: n + n + 1 + n + 2 = n + 3 Simplify by grouping like terms: (n + n + n) + (1 + 2) = n + 3 3n + 3 = n + 3 3n = n n = 0 n = 0 n + 1 = 1 n + 2 = 2 n + 3 = 3 Check our work: 0 + 1 +2 ? 3 3 = 3 Our final answer is [B](0, 1, 2, 3}[/B]

A baker makes 387 cupcakes. They are sold in packs of six. How many full packs can be made? How many cupcakes are leftover? Full packs = Lowest Rounded Integer of (Total cupcakes / packs) Full packs = Lowest Rounded Integer of 387/6 Full Packs = Lowest Rounded Integer of 64.5 Full Packs = [B]64[/B] Leftover = [URL='https://www.mathcelebrity.com/modulus.php?num=387mod6&pl=Calculate+Modulus']387 mod 6[/URL] Leftover = [B]3[/B]

A bank charges a service fee of $7.50 per month for a checking account. A bank account has $85.00. If no money is deposited or withdrawn except the service charge, how many months until the account balance is negative? Let m be the number of months. Our balance is denoted by B(m): B(m) = 85 - 7.5m The question asks when B(m) is less than 0. So we set up an inequality: 85 - 7.5m < 0 To solve this inequality for m, [URL='https://www.mathcelebrity.com/1unk.php?num=85-7.5m%3C0&pl=Solve']we type it in our search engine[/URL] and we get: m > 11.3333 We round up to the next whole integer and get [B]m = 12[/B]

a collection of 7 pencils, every week 3 more pencils are added How many weeks will it take to have 30 pencils? Set up a function, P(w), where w is the number of weeks, and P(w) is the total amount of pencils after w weeks. We have: P(w) = 3w + 7 We want to know what w is when P(w) = 30 3w + 7 = 30 [URL='https://www.mathcelebrity.com/1unk.php?num=3w%2B7%3D30&pl=Solve']Typing this equation into our search engine[/URL], we get: w = 7.6667 We round up to the nearest integer, so we get [B]w = 8[/B]

A Fahrenheit thermometer shows that the temperature is 15 degrees below zero. Enter the integer that represents the temperature in degrees Fahrenheit. Below zero means negative in Fahrenheit, so we have: [B]-15[/B]

A football team lost 7 yards each play for four consecutive plays. Represent the team’s total change in position for the four plays as an integer. A net loss in yardage for 7 yards is written as -7 4 plays * -7 yards equals [B]-28[/B]

A is the set of integers greater than or equal to -5 and less than or equal to -2 [B]-5 <= A <= -2[/B]

A is the set of odd integers between 4 and 12 Let A be the set of odd numbers between 4 and 12: [B]A = {5, 7, 9, 11}[/B]

A population grows at 6% per year. How many years does it take to triple in size? With a starting population of P, and triple in size means 3 times the original, we want to know t for: P(1.06)^t = 3P Divide each side by P, and we have: 1.06^t = 3 Typing this equation into our search engine to solve for t, we get: t = [B]18.85 years[/B] Note: if you need an integer answer, we round up to 19 years

A promotional deal for long distance phone service charges a $15 basic fee plus $0.05 per minute for all calls. If Joe's phone bill was $60 under this promotional deal, how many minutes of phone calls did he make? Round to the nearest integer if necessary. Let m be the number of minutes Joe used. We have a cost function of: C(m) = 0.05m + 15 If C(m) = 60, then we have: 0.05m + 15 = 60 [URL='https://www.mathcelebrity.com/1unk.php?num=0.05m%2B15%3D60&pl=Solve']Typing this equation into our search engine[/URL], we get: m = [B]900[/B]

A set of 4 consecutive integers adds up to 314. What is the least of the 4 integers? First integer is x. The next 3 are x + 1, x + 2, and x + 3. Set up our equation: x + (x + 1) + (x + 2) + (x + 3) = 314 Group x terms and group constnats (x + x + x + x) + (1 + 2 + 3) = 314 Simplify and combine 4x + 6 = 314 [URL='http://www.mathcelebrity.com/1unk.php?num=4x%2B6%3D314&pl=Solve']Enter this in the equation solver[/URL] [B]x = 77[/B]

A square of an integer is the integer. Find the integer. Let the integer be n. The square means we raise n to the power of 2, so we have: n^2 = n Subtract n from each side: n^2 - n = n - n n^2 - n = 0 Factoring this, we get: n(n - 1) = 0 So n is either [B]0 or 1[/B].

A submarine is 75 feet below sea level. It descends another 25 feet every 10 seconds for 3 minutes. What integer represents the submarines current location? Assumptions and givens: [LIST] [*]Let m be the number of minutes [*]10 seconds is 1/6 of a minute, 6 (10) seconds blocks per minute * 3 minutes = 18 (10 second blocks) [*]Below sea level is a negative number [/LIST] [U]Current depth:[/U] -25(18) - 75 -450 - 75 [B]-525[/B]

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Ages are consecutive integers. The sum of ages are 111. What are the ages In the search engine, we type [I][URL='http://www.mathcelebrity.com/consecintwp.php?num=111&pl=Sum']sum of 2 consecutive integers is 111[/URL][/I]. We get [B]55 and 56[/B].

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are all integers whole numbers true or false [B]False [/B] [LIST] [*]All whole numbers are integers but not all integers are whole numbers. [*]Whole numbers are positive integers. Which means negative integers are not whole numbers [*]-1 for instance is an integer, but not a whole number [/LIST]

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Before Barry Bonds, Mark McGwire, and Sammy Sosa, Roger Maris held the record for the most home runs in one season. Just behind Maris was Babe Ruth. The numbers of home runs hit by these two athletes in their record-breaking seasons form consecutive integers. Combined, the two athletes hit 121 home runs. Determine the number of home runs hit by Maris and Ruth in their record-breaking seasons. We want [URL='https://www.mathcelebrity.com/consecintwp.php?num=121&pl=Sum']the sum of 2 consecutive integers equals 121[/URL]. [B]We get Maris at 61 and Ruth at 60[/B]

Free Consecutive Integer Word Problems Calculator - Calculates the word problem for what two consecutive integers, if summed up or multiplied together, equal a number entered.

[INDENT]Consecutive odd integers are odd integers that differ by ______ , such as ______ and 13. Consecutive even integers are even integers that differ by ______ , such as 12 and ______ . Consecutive odd integers are odd integers that differ by ___2___ , such as ___11___ and 13. Consecutive even integers are even integers that differ by ___2___ , such as 12 and ___14___ .[/INDENT]

Crystal is serving pizza at a birthday party for her brother there are 25 people coming to the party she wants each each person to have 3 pieces of pizza each pizza has 8 slices how many pizzas should she buy? 25 people * 3 pieces of pizza each = 75 pieces of pizza Each pizza has 8 pieces. 75 pieces / 8 pieces per pizza = 9.375 pizzas. Round up to [B]10[/B] since we want an integer answer.

Dan needs 309 programs for the school play on Thursday. How many boxes of programs will he need, given that each box contains 41 programs? Each box contains 41 programs, so we divide 309 programs by 41 programs per box to get our boxes: 309/41 using our [URL='https://www.mathcelebrity.com/longdiv.php?num1=309&num2=41&pl=Long%20Division%20%28Decimals%29']division calculator[/URL] is 7.5365. Since we don't have fractional boxes, we round up to the next highest integer. [B]8 boxes[/B]

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Find 3 consecutive integers such that the sum of twice the smallest and 3 times the largest is 126. Let the first integer be n, the second integer be n + 1, and the third integer be n + 2. We have: Sum of the smallest and 3 times the largest is 126: n + 3(n + 2) = 126 Multiply through: n + 3n + 6 = 126 Group like terms: 4n + 6 = 126 [URL='https://www.mathcelebrity.com/1unk.php?num=4n%2B6%3D126&pl=Solve']Type 4n + 6 = 126 into our calculator[/URL], we get n = 30. Which means the next two integers are 31 and 32. [B]{30, 31, 32}[/B]

Find 3 Even Integers with a sum of 198 Let x be the first even integer. Then y is the next, and z is the third even integer. [LIST=1] [*]y = x + 2 [*]z = x + 4 [*]x + y + z = 198 [/LIST] Substituting y and z into (3): x + x + 2 + x + 4 = 198 Group x terms 3x + 6 = 198 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=3x%2B6%3D198&pl=Solve']equation solver[/URL], we get: [B]x = 64[/B] y = 64 + 2 [B]y= 66[/B] z = 64 + 4 [B]z = 68[/B]

Find the largest of three consecutive even integers when six times the first integers is equal to five times the middle integer. Let the first of the 3 consecutive even integers be n. The second consecutive even integer is n + 2. The third (largest) consecutive even integer is n + 4. We are given 6n = 5(n + 2). Multiply through on the right side, and we get: 6n = 5n + 10 [URL='https://www.mathcelebrity.com/1unk.php?num=6n%3D5n%2B10&pl=Solve']Typing 6n = 5n + 10 into the search engine[/URL], we get n = 10. Remember, n was our smallest of 3 consecutive even integers. So the largest is: n + 4 10 + 4 [B]14[/B]

Find two consecutive integers if the sum of their squares is 1513 Let the first integer be n. The next consecutive integer is (n + 1). The sum of their squares is: n^2 + (n + 1)^2 = 1513 n^2 + n^2 + 2n + 1 = 1513 2n^2 + 2n + 1 = 1513 Subtract 1513 from each side: 2n^2 + 2n - 1512 = 0 We have a quadratic equation. We [URL='https://www.mathcelebrity.com/quadratic.php?num=2n%5E2%2B2n-1512%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']type this into our search engine[/URL] and get: n = (-27, 28) Let's take the positive solution. The second integer is: n + 1 28 + 1 = 29

Find two consecutive intergers whose sum is 15 Use our [URL='http://www.mathcelebrity.com/consecintwp.php?pl=Sum&num=+15']consecutive integer calculator[/URL], we get: (7, 8)

Find two consecutive odd integers such that the sum of their squares is 290. Let the first odd integer be n. The next odd integer is n + 2 Square them both: n^2 (n + 2)^2 = n^2 + 4n + 4 from our [URL='https://www.mathcelebrity.com/expand.php?term1=%28n%2B2%29%5E2&pl=Expand']expansion calculator[/URL] The sum of the squares equals 290 n^2 + n^2 + 4n + 4 = 290 Group like terms: 2n^2 + 4n + 4 = 290 [URL='https://www.mathcelebrity.com/quadratic.php?num=2n%5E2%2B4n%2B4%3D290&pl=Solve+Quadratic+Equation&hintnum=+0']Enter this quadratic into our search engine[/URL], and we get: n = 11, n = -13 Which means the two consecutive odd integer are: 11 and 11 + 2 = 13. [B](11, 13)[/B] -13 and -13 + 2 = -11 [B](-13, -11)[/B]

Find two consecutive positive integers such that the difference of their square is 25. Let the first integer be n. This means the next integer is (n + 1). Square n: n^2 Square the next consecutive integer: (n + 1)^2 = n^2 + 2n + 1 Now, we take the difference of their squares and set it equal to 25: (n^2 + 2n + 1) - n^2 = 25 Cancelling the n^2, we get: 2n + 1 = 25 [URL='https://www.mathcelebrity.com/1unk.php?num=2n%2B1%3D25&pl=Solve']Typing this equation into our search engine[/URL], we get: n = [B]12[/B]

Find two consecutive positive integers such that the sum of their squares is 25. Let the first integer be x. The next consecutive positive integer is x + 1. The sum of their squares equals 25. We write this as:: x^2 + (x + 1)^2 Expanding, we get: x^2 + x^2 + 2x + 1 = 25 Group like terms: 2x^2 + 2x + 1 = 25 Subtract 25 from each side: 2x^2 + 2x - 24 = 0 Simplify by dividing each side by 2: x^2 + x - 12 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-12%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get x = 3 or x = -4. The problem asks for positive integers, so we discard -4, and use 3. This means, our next positive integer is 3 + 1 = 4. So we have [B](3, 4) [/B]as our answers. Let's check our work: 3^2 + 4^2 = 9 + 16 = 25

Flight is $295 and car rental is $39 a day, if a competition charges $320 and $33 a day car rental, which is cheaper? Set up cost function where d is the number of days: [LIST] [*]Control business: C(d) = 39d + 295 [*]Competitor business: C(d) = 33d + 320 [/LIST] Set the [URL='http://www.mathcelebrity.com/1unk.php?num=39d%2B295%3D33d%2B320&pl=Solve']cost functions equal to each other[/URL]: We get d = 4.1667. The next integer day up is 5. Now plug in d = 1, 2, 3, 4. For the first 4 days, the control business is cheaper. However, starting at day 5, the competitor business is now cheaper forever.

Four consecutive integers beginning with n consecutive meaning one after another. So we have: [LIST] [*][B]n[/B] [*][B]n + 1[/B] [*][B]n + 2[/B] [*][B]n + 3[/B] [/LIST]

Four cousins were born at two-year intervals. The sum of their ages is 36. What are their ages? So the last cousin is n years old. this means consecutive cousins are n + 2 years older than the next. whether their ages are even or odd, we have the sum of 4 consecutive (odd|even) integers equal to 36. We [URL='https://www.mathcelebrity.com/sum-of-consecutive-numbers.php?num=sumof4consecutiveevenintegersis36&pl=Calculate']type this into our search engine[/URL] and we get the ages of: [B]6, 8, 10, 12[/B]

Yes, $40 for both. This was my son's 8th grade problem. They are learning integers with special cases, so this makes sense, I hope. Thank you so much for responding. I did have this written, but I wasn't certain it was correct.

If 2 times an integer x is increased by 5 2 times an integer x: 2x The phrase [I]increased by[/I] means we add, so we add 5 to 2x: [B]2x + 5[/B]

if 2z-1 is an odd integer what is the preceding odd integer? The preceding odd integer is found by subtracting 2: 2z - 1 - 2 [B]2z - 3[/B]

If 7 times the square of an integer is added to 5 times the integer, the result is 2. Find the integer. [LIST] [*]Let the integer be "x". [*]Square the integer: x^2 [*]7 times the square: 7x^2 [*]5 times the integer: 5x [*]Add them together: 7x^2 + 5x [*][I]The result is[/I] means an equation, so we set 7x^2 + 5x equal to 2 [/LIST] 7x^2 + 5x = 2 [U]This is a quadratic equation. To get it into standard form, we subtract 2 from each side:[/U] 7x^2 + 5x - 2 = 2 - 2 7x^2 + 5x - 2 = 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=7x%5E2%2B5x-2%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']Type this problem into our search engine[/URL], and we get two solutions: [LIST=1] [*]x = 2/7 [*]x= -1 [/LIST] The problem asks for an integer, so our answer is x[B] = -1[/B]. [U]Let's check our work by plugging x = -1 into the quadratic:[/U] 7x^2 + 5x - 2 = 0 7(-1)^2 + 5(-1) - 2 ? 0 7(1) - 5 - 2 ? 0 0 = 0 So we verified our answer, [B]x = -1[/B].

if a and b are odd then a + b is even Let a and b be positive odd integers of the form: [LIST] [*]a = 2n + 1 [*]b = 2m + 1 [/LIST] a + b = 2n + 1 + 2m + 1 a + b = 2n + 2m + 1 + 1 Combing like terms, we get: a + b = 2n + 2m + 2 a + b = 2(n + m) + 2 Let k = n + m a + b = 2k + 2 [B]Therefore a + b is even[/B]

if a divides b, then a divides bc Suppose a divides b. Then there exists an integer q such that b = aq, so that bc = a(qc) and a divides bc. Suppose that a divides c. Then there exists an integer k such that c = ak, so that bc = a(kb) and a divides bc.

If a is an even integer and b is an odd integer then prove a − b is an odd integer Let a be our even integer Let b be our odd integer We can express a = 2x (Standard form for even numbers) for some integer x We can express b = 2y + 1 (Standard form for odd numbers) for some integer y a - b = 2x - (2y + 1) a - b = 2x - 2y - 1 Factor our a 2 from the first two terms: a - b = 2(x - y) - 1 Since x - y is an integer, 2(x- y) is always even. Subtracting 1 makes this an odd number. [MEDIA=youtube]GDVuQ7bGHx8[/MEDIA]

If a, b, and c are positive integers such that a^b = x and c^b = y, then xy = ? A) ac^b B) ac^2b C) (ac)^b D) (ac)^2b E) (ac)^b^2 xy = a^b * c^b We can use the Power of a Product Rule a^b * c^b = (ac)^b Therefore: xy = [B](ac)^b - Answer C[/B]

If first integer is 5y, then the next two consecutive integers are integers increase by 1, so we have: [B]5y + 1[/B] 5y + 1 + 1 = [B]5y + 2[/B]

Look at the Contrapositive: If n is even, then 3n + 2 is even... Suppose that the conclusion is false, i.e., that n is even. Then n = 2k for some integer k. Then we have: 3n + 2 = 3(2k) + 2 3n + 2 = 6k + 2 3n + 2 = 2(3k + 1). Thus 3n + 2 is even, because it equals 2j for an integer j = 3k + 1. So 3n + 2 is not odd. We have shown that ¬(n is odd) → ¬(3n + 2 is odd), therefore, the contrapositive (3n + 2 is odd) → (n is odd) is also true.

If n represents an odd integer what represents the previous smaller odd integer Each odd integer is 2 away from the last one, so the previous smaller odd integer is found by subtracting 2 from n: [B]n - 2[/B]

If x is divided by 9, the remainder is 5. What is the remainder if 3x is divided by 9? pick an integer x where when dividing by 9, we get a remainder of 5. 14/9 gives us a remainder of 5. Now multiply 14 by 3: 14 * 3 = 42 [URL='https://www.mathcelebrity.com/modulus.php?num=42mod9&pl=Calculate+Modulus']42/9 gives a remainder of[/URL] [B]6[/B]

If x represents the first, or the smaller, of two consecutive odd integers, express the sum of the two integers in terms of x If x is the first of two consecutive odd integers, then we find the next consecutive odd integer by adding 2 to x: x + 2 The sum of the two consecutive odd integers is expressed by x + (x + 2) Simplify by grouping like terms, we get: [B]2x + 2[/B]

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Let n be an integer. If n^2 is odd, then n is odd Proof by contraposition: Suppose that n is even. Then we can write n = 2k n^2 = (2k)^2 = 4k^2 = 2(2k) so it is even [I]So an odd number can't be the square of an even number. So if an odd number is a square it must be the square of an odd number.[/I]

Let n be the middle number of three consecutive integers This means: [LIST] [*]n is the second of three consecutive integers [*]The first consecutive integer is n - 1 [*]The third consecutive integer is n + 1 [/LIST] The sum is found by: n - 1 + n + n + 1 Simplifying, we get: (n + n + n) + 1 - 1 [B]3n[/B]

Let P(n) and S(n) denote the product and the sum, respectively, of the digits of the integer n. For example, P(23) = 6 and S(23) = 5. Suppose N is a two-digit number such that N = P(N) + S(N). What could N be? Is there more than one answer? For example, for 23 P(23) = 6 and S(23) = 5, but 23 could not be the N that we want since 23 <> 5 + 6 Let t = tens digit and o = ones digit P(n) = to S(n) = t + o P(n) + S(n) = to + t + o N = 10t + o Set them equal to each other N = P(N) + S(N) 10t + o = to + t + o o's cancel, so we have 10t = to + t Subtract t from each side, we have 9t = to Divide each side by t o = 9 So any two-digit number with 9 as the ones digit will work: [B]{19,29,39,49,59,69,79,89,99}[/B]

Let U be the set of all integers between −3 and 3 (including −3 and 3). Let A={−2,0,1,3}. Find Ac. Give your answer in standard set notation Ac is anything not in A, but in U. So we have: Ac = [B]{-3, -1, 2}[/B]

Let x be an integer. If x is odd, then x^2 is odd Proof: Let x be an odd number. This means that x = 2n + 1 where n is an integer. [U]Squaring x, we get:[/U] x^2 = (2n + 1)^2 = (2n + 1)(2n + 1) x^2 = 4n^2 + 4n + 1 x^2 = 2(2n^2 + 2n) + 1 2(2n^2 + 2n) is an even number since 2 multiplied by any integer is even So adding 1 is an odd number [MEDIA=youtube]GlzV80M33x0[/MEDIA]

M is the set of integers that are greater than or equal to -1 and less than or equal to 2 We include -1 on the left, and include 2 on the right [B]M = {-1, 0, 1, 1, 2)[/B]

Free Modulus Calculator - Given 2 integers a and b, this modulo calculator determines a mod b or simplifies modular arithmetic such as 7 mod 3 + 5 mod 8 - 32 mod 5

natural numbers that are factors of 16 Natural numbers are positive integers starting at 1. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16} Of these, [URL='https://www.mathcelebrity.com/factoriz.php?num=16&pl=Show+Factorization']the only factors of 16[/URL] are: {[B]1, 2, 4, 8, 16}[/B]

What do you want to do with this number set? Express it as representation of integers?

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Free Number Property Calculator - This calculator determines if an integer you entered has any of the following properties:
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Free Opposite Numbers Calculator - Given a positive or negative integer (n), this calculates the opposite number of n

Free Ordinal Number Calculator - This calculator determines the ordinal number of an integer

Update, this has been added to our shortcuts. You can type any expression in the form, positive numbers less than x where x is any integer. You can also type positive numbers greater than x where x is any integer. Same with less than or equal to and greater than or equal to.

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Prove 0! = 1 Let n be a whole number, where n! represents the product of n and all integers below it through 1. The factorial formula for n is: n! = n · (n - 1) * (n - 2) * ... * 3 * 2 * 1 Written in partially expanded form, n! is: n! = n * (n - 1)! [U]Substitute n = 1 into this expression:[/U] n! = n * (n - 1)! 1! = 1 * (1 - 1)! 1! = 1 * (0)! For the expression to be true, 0! [U]must[/U] equal 1. Otherwise, 1! <> 1 which contradicts the equation above

[URL='https://www.mathcelebrity.com/proofs.php?num=prove0%21%3D1&pl=Prove']Prove 0! = 1[/URL] Let n be a whole number, where n! represents: The product of n and all integers below it through 1. The factorial formula for n is n! = n · (n - 1) · (n - 2) · ... · 3 · 2 · 1 Written in partially expanded form, n! is: n! = n · (n - 1)! [SIZE=5][B]Substitute n = 1 into this expression:[/B][/SIZE] n! = n · (n - 1)! 1! = 1 · (1 - 1)! 1! = 1 · (0)! For the expression to be true, 0! [U]must[/U] equal 1. Otherwise, 1! ≠ 1 which contradicts the equation above [MEDIA=youtube]wDgRgfj1cIs[/MEDIA]

Use proof by contradiction. Assume sqrt(2) is rational. This means that sqrt(2) = p/q for some integers p and q, with q <>0. We assume p and q are in lowest terms. Square both side and we get: 2 = p^2/q^2 p^2 = 2q^2 This means p^2 must be an even number which means p is also even since the square of an odd number is odd. So we have p = 2k for some integer k. From this, it follows that: 2q^2 = p^2 = (2k)^2 = 4k^2 2q^2 = 4k^2 q^2 = 2k^2 q^2 is also even, therefore q must be even. So both p and q are even. This contradicts are assumption that p and q were in lowest terms. So sqrt(2) [B]cannot be rational. [MEDIA=youtube]tXoo9-8Ewq8[/MEDIA][/B]

Take an integer n. The next alternate consecutive integer is n + 2 Subtract the difference of the squares: (n + 2)^2 - n^2 n^2 + 4n + 4 - n^2 n^2 terms cancel, we get: 4n + 4 Factor out a 4: 4(n + 1) If n is odd, n + 1 is even. 4 * even is always even If n is even, n + 1 is odd. 4 * odd is always odd Since both cases are even, we've proven our statement. [MEDIA=youtube]J_E9lR5qFY0[/MEDIA]

Take two consecutive integers: n, n + 1 The difference of their cubes is: (n + 1)^3 - n^3 n^3 + 3n^2 + 3n + 1 - n^3 Cancel the n^3 3n^2 + 3n + 1 Factor out a 3 from the first 2 terms: 3(n^2 + n) + 1 The first two terms are always divisible by 3 but then the + 1 makes this expression not divisible by 3: 3(n^2 + n) + 1 = 1 (mod 3) [MEDIA=youtube]hFvJ3epqmyE[/MEDIA]

Take an integer n. The next consecutive integer is n + 1 Subtract the difference of the squares: (n + 1)^2 - n^2 n^2 + 2n + 1 - n^2 n^2 terms cancel, we get: 2n + 1 2 is even. For n, if we use an even: we have even * even = Even Add 1 we have Odd 2 is even. For n, if we use an odd: we have even * odd = Even Add 1 we have Odd Since both cases are odd, we've proven our statement. [MEDIA=youtube]RAi0HbH5bqc[/MEDIA]

Prove the following statement for non-zero integers a, b, c, If a divides b and b divides c, then a divides c. If an integer a divides an integer b, then we have: b = ax for some non-zero integer x If an integer b divides an integer c, then we have: c = by for some non-zero integer y Since b = ax, we substitute this into c = by for b: c = axy We can write this as: c = a(xy) [LIST] [*]Since x and y are integers, then xy is also an integer. [*]Therefore, c is the product of some integer multiplied by a [*]This means a divides c [/LIST] [MEDIA=youtube]VUIUFAFFVU4[/MEDIA]

Take two integers, r and s. We can write r as a/b for integers a and b since a rational number can be written as a quotient of integers We can write s as c/d for integers c and d since a rational number can be written as a quotient of integers Add r and s: r + s = a/b + c/d With a common denominator bd, we have: r + s = (ad + bc)/bd Because a, b, c, and d are integers, ad + bc is an integer since rational numbers are closed under addition and multiplication. Since b and d are non-zero integers, bd is a non-zero integer. Since we have the quotient of 2 integers, r + s is a rational number. [MEDIA=youtube]0ugZSICt_bQ[/MEDIA]

Take two arbitrary integers, x and y We can express the odd integer x as 2a + 1 for some integer a We can express the odd integer y as 2b + 1 for some integer b x + y = 2a + 1 + 2b + 1 x + y = 2a + 2b + 2 Factor out a 2: x + y = 2(a + b + 1) Since 2 times any integer even or odd is always even, then [B]x + y by definition is even[/B]. [MEDIA=youtube]9A-qe4yZXYw[/MEDIA]

Let us take an integer x which is both even [I]and[/I] odd. [LIST] [*]As an even integer, we write x in the form 2m for some integer m [*]As an odd integer, we write x in the form 2n + 1 for some integer n [/LIST] Since both the even and odd integers are the same number, we set them equal to each other 2m = 2n + 1 Subtract 2n from each side: 2m - 2n = 1 Factor out a 2 on the left side: 2(m - n) = 1 By definition of divisibility, this means that 2 divides 1. But we know that the only two numbers which divide 1 are 1 and -1. Therefore, our original assumption that x was both even and odd must be false. [MEDIA=youtube]SMM9ubEVcLE[/MEDIA]

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Rick sold a total of 75 books during the first 22 days of May. If he continues to sell books at the same rate, how many books will he sell during the month of May? Set up a proportion of days to books where n is the number of books sold in May: 22/31 = 75/n Using our [URL='https://www.mathcelebrity.com/proportion-calculator.php?num1=22&num2=75&den1=31&den2=n&propsign=%3D&pl=Calculate+missing+proportion+value']proportion calculator[/URL] and rounding to the next integer, we get: n = [B]106[/B]

Free Roman Numeral Conversions Calculator - Converts a Positive integer less than 4000 to a Roman Numeral.
Converts a Roman Numeral with a positive value less than 4000 to a number.

Roster form of: A = {3x-2/x are integers between 0 and 8} x = 0 = Undefined since we divide by 0 x = 1: 3*1 + 2/1 = 5 x = 2: 3*2 + 2/2 = 7 x = 3: 3*3 + 2/3 = 9.66666666666667 x = 4: 3*4 + 2/4 = 12.5 x = 5: 3*5 + 2/5 = 15.4 x = 6: 3*6 + 2/6 = 18.3333333333333 x = 7: 3*7 + 2/7 = 21.2857142857143 x = 8: 3*8 + 2/8 = 24.25 [B]A = {(0, undefined), (1, 5), (2, 7), (3, 9.6667), (4, 12.5), (5, 15.4), (6, 18.3333), (7, 21.2857142857143), (8, 24.25)}[/B]

Sam and Jeremy have ages that are consecutive odd integers. The product of their ages is 783. Which equation could be used to find Jeremy's age, j, if he is the younger man. Let Sam's age be s. Let' Jeremy's age be j. We're given: [LIST=1] [*]s = j + 2 <-- consecutive odd integers [*]sj = 783 [/LIST] Substitute (1) into (2): (j + 2)j = 783 j^2 + 2j = 783 Subtract 783 from each side: j^2 + 2j - 783 = 0 <-- This is the equation to find Jeremy's age. To solve this, [URL='https://www.mathcelebrity.com/quadratic.php?num=j%5E2%2B2j-783%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']we type this quadratic equation into the search engine[/URL] and get: j = 27, j = -29. Since ages cannot be negative, we have: [B]j = 27[/B]

Free Signed Integer Operations Calculator - This performs a string of signed integer operations, either all addition and subtraction, or all multiplication and division.

Standard Error (margin of Error) = Standard Deviation / sqrt(n) 128 = 545/sqrt(n) Cross multiply: 128sqrt(n) = 545 Divide by 128 sqrt(n) = 4.2578125 Square both sides: [B]n = 18.1289672852 But we need an integer, so the answer is 19[/B]

Storm Center 7 was tracking a cold front approaching Cedarburg. Before it rolled in, the temperature was – 7°F. Then, the temperature decreased by 9°F. What was the temperature after the cold front rolled in? Using signed integers, we start with 7 below or -7 -7 The temperature decreased by 9 which means we subtract: -7 - 9 or -7 + (-9) [B]-16°F or 16 below 0 [MEDIA=youtube]oJjEhkdnTxA[/MEDIA][/B]

sum of 3 consecutive odd integers equals 1 hundred 17 The sum of 3 consecutive odd numbers equals 117. What are the 3 odd numbers? 1) Set up an equation where our [I]odd numbers[/I] are n, n + 2, n + 4 2) We increment by 2 for each number since we have [I]odd numbers[/I]. 3) We set this sum of consecutive [I]odd numbers[/I] equal to 117 n + (n + 2) + (n + 4) = 117 [SIZE=5][B]Simplify this equation by grouping variables and constants together:[/B][/SIZE] (n + n + n) + 2 + 4 = 117 3n + 6 = 117 [SIZE=5][B]Subtract 6 from each side to isolate 3n:[/B][/SIZE] 3n + 6 - 6 = 117 - 6 [SIZE=5][B]Cancel the 6 on the left side and we get:[/B][/SIZE] 3n + [S]6[/S] - [S]6[/S] = 117 - 6 3n = 111 [SIZE=5][B]Divide each side of the equation by 3 to isolate n:[/B][/SIZE] 3n/3 = 111/3 [SIZE=5][B]Cancel the 3 on the left side:[/B][/SIZE] [S]3[/S]n/[S]3 [/S]= 111/3 n = 37 Call this n1, so we find our other 2 numbers n2 = n1 + 2 n2 = 37 + 2 n2 = 39 n3 = n2 + 2 n3 = 39 + 2 n3 = 41 [SIZE=5][B]List out the 3 consecutive odd numbers[/B][/SIZE] ([B]37, 39, 41[/B]) 37 ← 1st number, or the Smallest, Minimum, Least Value 39 ← 2nd number 41 ← 3rd or the Largest, Maximum, Highest Value

Free Sum of Consecutive Numbers Calculator - Finds the sum of (n) consecutive integers, even or odd as well. Examples include:
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Free Sum of Five Consecutive Integers Calculator - Finds five consecutive integers, if applicable, who have a sum equal to a number. Sum of 5 consecutive integers

Free Sum of Four Consecutive Integers Calculator - Finds four consecutive integers, if applicable, who have a sum equal to a number. Sum of 4 consecutive integers

Sum of N and its next consecutive even integer is 65 Next even consecutive integer is N + 2. We have N + (N + 2) = 65. Combine like terms, we have 2N + 2 = 65 [URL='http://www.mathcelebrity.com/1unk.php?num=2n%2B2%3D65&pl=Solve']Running this problem through the search engine[/URL], we get n = 31.5. Meaning this problem is impossible, it cannot be done. n is not an integer, and neither is the next consecutive even integer.

Free Sum of Three Consecutive Integers Calculator - Finds three consecutive integers, if applicable, who have a sum equal to a number. Sum of 3 consecutive integers

Sum of two consecutive numbers is always odd Definition: [LIST] [*]A number which can be written in the form of 2 m where m is an integer, is called an even integer. [*]A number which can be written in the form of 2 m + 1 where m is an integer, is called an odd integer. [/LIST] Take two consecutive integers, one even, and one odd: 2n and 2n + 1 Now add them 2n + (2n+ 1) = 4n + 1 = 2(2 n) + 1 The sum is of the form 2n + 1 (2n is an integer because the product of two integers is an integer) Therefore, the sum of two consecutive integers is an odd number.

Free Tally Marks Calculator - Shows the tally mark representation (tallies) for a positive integer.

The ages of three siblings are all consecutive integers. The sum of of their ages is 39. Let the age of the youngest sibling be n. This means the second sibling is n + 1. This means the oldest/third sibling is n + 2. So what we want is the[URL='https://www.mathcelebrity.com/sum-of-consecutive-numbers.php?num=sumof3consecutiveintegersequalto39&pl=Calculate'] sum of 3 consecutive integers equal to 39[/URL]. We type this command into our search engine. We get: n = 12. So the youngest sibling is [B]12[/B]. The next sibling is 12 + 1 = [B]13[/B] The oldest/third sibling is 12 + 2 = [B]14[/B]

The cost of renting a rototiller is $19.50 for the first hour and $7.95 for each additional hour. How long can a person have the rototiller if the cost must be less than $95? Setup the inequality: $19.50 + $7.95x < $95 Subtract 19.50 from both sides: 7.95x < 75.50 Divide each side of the inequality by 7.95 to isolate x x < 9.5 The next lowest integer is 9. So we take 9 + the first hour of renting to get [B]10 total hours[/B]. Check our work: $7.95 * 9.5 + $19.50 $71.55 + $19.50 = $91.05

The domain of a relation is all even negative integers greater than -9. The range y of the relation is the set formed by adding 4 to the numbers in the domain. Write the relation as a table of values and as an equation. The domain is even negative integers greater than -9: {-8, -6, -4, -2} Add 4 to each x for the range: {-8 + 4 = -4, -6 + 4 = -2. -4 + 4 = 0, -2 + 4 = 2} For ordered pairs, we have: (-8, -4) (-6, -2) (-4, 0) (-2, 2) The equation can be written: y = x + 4 on the domain (x | x is an even number where -8 <= x <= -2)

The expression (5x - 2)/(x + 3) is equivalent to which of the following? [LIST] [*]A) (5 - 2)/3 [*]B) 5 - 2/3 [*]C) 5 - (2/(x + 3)) [*]D) 5 - (17/(x + 3)) [/LIST] Let's start with an integer x = 2. Plug that into our original expression, and we get: (5(2) - 2)(2 + 3) (10 - 2)/5 8/5 So what we do next is, take x = 2, and plug it into answer choices A-D, and see which one results in 8/5 A) 3/3 = 1 <-- Nope B) Since 5 is 15/3, we have 15/3 - 2/3 = 13/3 which is over 4, so Nope C) 5 - (2/(2 + 3)) = 5 - (2/5). Since 5 is 25/5, we have 25/5 - 2/5 = 23/5. <-- Nope D) 5 - (17/(2 + 3)) = 5 - 17/5. Since 5 is 25/5, we have 25/5 - 17/5 = 8/5 <-- YES Since 8/5 = 8/5, our answer is [B]D) 5 - (17/(x + 3))[/B]

the left and right page numbers of an open book are two consecutive integers whose number is 235 find the page numbers Using our [URL='https://www.mathcelebrity.com/consecintwp.php?pl=Sum&num=+235']consecutive integer calculator[/URL], we get: [B]117, 118[/B]

The left and right page numbers of an open book are two consecutive integers whose sum is 403. Find these page numbers. Page numbers left and right are consecutive integers. So we want to find a number n and n + 1 where: n + n + 1 = 403 Combining like terms, we get: 2n + 1 = 403 Typing that equation into our search engine, we get: [B]n = 201[/B] This is our left hand page. Our right hand page is: 201 + 1 = [B]202[/B]

The number -2.34 can be found between which two integers We want to take the integer of -2.34 which is -2 Since -2.34 is less than 0, we subtract 1: -2 -1 = -3 Therefore, -2.34 lies between [B]-2 and -3[/B] -3 < -2.34 < -2

The product of two consecutive integers is greater than 100 Take an integer x. Next consecutive integer is x + 1 The product of those integers is: x(x + 1) This product is greater than 100 which gives us the algebraic expression of: x(x + 1) > 100 IF we want to solve for x: x^2 + x > 100 Subtract 100 from each side: x^2 + x - 100 > 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-100%3E0&pl=All&hintnum=+0']Solve this quadratic:[/URL]

The Square of a positive integer is equal to the sum of the integer and 12. Find the integer Let the integer be x. [LIST] [*]The sum of the integer and 12 is written as x + 12. [*]The square of a positive integer is written as x^2. [/LIST] We set these equal to each other: x^2 = x + 12 Subtract x + 12 from each side: x^2 - x - 12 = 0 We have a quadratic function. [URL='https://www.mathcelebrity.com/quadratic.php?num=x%5E2-x-12%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']Run it through our search engine[/URL] and we get x = 3 and x = -4. The problem asks for a positive integer, so we have [B]x = 3[/B]

The square of a positive integer minus twice its consecutive integer is equal to 22. Find the integers. Let x = the original positive integer. We have: [LIST] [*]Consecutive integer is x + 1 [*]x^2 - 2(x + 1) = 22 [/LIST] Multiply through: x^2 - 2x - 2 = 22 Subtract 22 from each side: x^2 - 2x - 24 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2-2x-24%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get: x = 6 and x = -4 Since the problem states [U]positive integers[/U], we use: x = 6 and x + 1 = 7 [B](6, 7)[/B]

The sum of 3 consecutive integers is greater than 30. Let the first consecutive integer be n The second consecutive integer is n + 1 The third consecutive integer is n + 2 The sum is written as: n + n + 1 + n + 2 Combine like terms: (n + n + n) + (1 + 2) 3n + 3 The phrase [I]greater than[/I] means an inequality, which we write as: [B]3n + 3 > 30[/B]

The sum of the squares of two consecutive positive integers is 61. Find these two numbers. Let the 2 consecutive integers be x and x + 1. We have: x^2 + (x + 1)^2 = 61 Simplify: x^2 + x^2 + 2x + 1 = 61 2x^2 + 2x + 1 = 61 Subtract 61 from each side: 2x^2 + 2x - 60 = 0 Divide each side by 2 x^2 + x - 30 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-30&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic equation calculator[/URL], we get: x = 5 and x = -6 The question asks for [I]positive integers[/I], so we use [B]x = 5. [/B]This means the other number is [B]6[/B].

Let the 3 integers be x, y, and z. y = x + 1 z = y + 1, or x + 2. Set up our equation: x + (x + 1) + (x + 2) = 42 Group our variables and constants: (x + x + x) + (1 + 2) = 42 3x + 3 = 42 Subtract 3 from each side: 3x = 39 Divide each side of the equation by 3: [B]x = 13 So y = x + 1 = 14 z = x + 2 = 15 (x,y,z) = (13,14,15)[/B]

The sum of twice an integer and 3 times the next consecutive integer is 48 Let the first integer be n This means the next consecutive integer is n + 1 Twice an integer means we multiply n by 2: 2n 3 times the next consecutive integer means we multiply (n + 1) by 3 3(n + 1) The sum of these is: 2n + 3(n + 1) The word [I]is[/I] means equal to, so we set 2n + 3(n + 1) equal to 48: 2n + 3(n + 1) = 48 Solve for [I]n[/I] in the equation 2n + 3(n + 1) = 48 We first need to simplify the expression removing parentheses Simplify 3(n + 1): Distribute the 3 to each term in (n+1) 3 * n = (3 * 1)n = 3n 3 * 1 = (3 * 1) = 3 Our Total expanded term is 3n + 3 Our updated term to work with is 2n + 3n + 3 = 48 We first need to simplify the expression removing parentheses Our updated term to work with is 2n + 3n + 3 = 48 [SIZE=5][B]Step 1: Group the n terms on the left hand side:[/B][/SIZE] (2 + 3)n = 5n [SIZE=5][B]Step 2: Form modified equation[/B][/SIZE] 5n + 3 = + 48 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 3 and 48. To do that, we subtract 3 from both sides 5n + 3 - 3 = 48 - 3 [SIZE=5][B]Step 4: Cancel 3 on the left side:[/B][/SIZE] 5n = 45 [SIZE=5][B]Step 5: Divide each side of the equation by 5[/B][/SIZE] 5n/5 = 45/5 Cancel the 5's on the left side and we get: n = [B]9[/B]

The sum of two consecutive integers if n is the first integer. consecutive means immediately after, so we have: n n + 1 [U]The sum is written as:[/U] n + n + 1 [U]Grouping like terms, we have:[/U] (n + n) + 1 [B]2n + 1[/B]

The sum of two consecutive integers plus 18 is 123. Let our first integer be n and our next integer be n + 1. We have: n + (n + 1) + 18 = 123 Group like terms to get our algebraic expression: 2n + 19 = 123 If we want to solve the algebraic expression using our [URL='http://www.mathcelebrity.com/1unk.php?num=2n%2B19%3D123&pl=Solve']equation solver[/URL], we get n = 52. This means the next integer is 52 + 1 = 53

There are 2 consecutive integers. Twice the first increased by the second yields 16. What are the numbers? Let x be the first integer. y = x + 1 is the next integer. We have the following givens: [LIST=1] [*]2x + y = 16 [*]y = x + 1 [/LIST] Substitute (2) into (1) 2x + (x + 1) = 16 Combine x terms 3x + 1 = 16 Subtract 1 from each side 3x = 15 Divide each side by 3 [B]x = 5[/B] So the other integer is 5 + 1 = [B]6[/B]

Three good friends are in the same algebra class, their scores on a recent test are three consecutive odd integers whose sum is 273. Find the score In our search engine, we type in [URL='https://www.mathcelebrity.com/sum-of-consecutive-numbers.php?num=3consecutiveintegerswhosesumis273&pl=Calculate']3 consecutive integers whose sum is 273[/URL] and we get: [B]90, 91, 92[/B]

To make an international telephone call, you need the code for the country you are calling. The code for country A, country B, and C are three consecutive integers whose sum is 90. Find the code for each country. If they are three consecutive integers, then we have: [LIST=1] [*]B = A + 1 [*]C = B + 1, which means C = A + 2 [*]A + B + C = 90 [/LIST] Substitute (1) and (2) into (3) A + (A + 1) + (A + 2) = 90 Combine like terms 3A + 3 = 90 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=3a%2B3%3D90&pl=Solve']equation calculator[/URL], we get: [B]A = 29[/B] Which means: [LIST] [*]B = A + 1 [*]B = 29 + 1 [*][B]B = 30[/B] [*]C = A + 2 [*]C = 29 + 2 [*][B]C = 31[/B] [/LIST] So we have [B](A, B, C) = (29, 30, 31)[/B]

Today is my birthday! Four-fifths of my current age is greater than three-quarters of my age one year from now. Given that my age is an integer number of years, what is the smallest my age could be? Let my current age be a. We're given: 4/5a > 3/4(a + 1) Multiply through on the right side: 4a/5 > 3a/4 + 3/4 Let's remove fractions by multiply through by 5: 5(4a/5) > 5(3a/4) + 5(3/4) 4a > 15a/4 + 15/4 Now let's remove the other fractions by multiply through by 4: 4(4a) > 4(15a/4) + 4(15/4) 16a > 15a + 15 [URL='https://www.mathcelebrity.com/1unk.php?num=16a%3E15a%2B15&pl=Solve']Typing this inequality into our search engine[/URL], we get: a > 15 This means the smallest [I]integer age[/I] which the problem asks for is: 15 + 1 = [B]16[/B]

Two consecutive even integers that equal 126 Let the first integer equal x. So the next even integer must be x + 2. The sum which is equal to 126 is written as x + (x + 2) = 126 Simplify: 2x + 2 = 126 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=2x%2B2%3D126&pl=Solve']equation calculator,[/URL] we get: x = 62 This means the next consecutive even integer is 62 = 2 = 64. So our two even consecutive integers with a sum of 126 are [B](62, 64)[/B]

what integer is tripled when 9 is added to 3 fourths of it? Let the integer be n. Tripling an integer means multiplying it by 3. We're given: 3n = 3n/4 + 9 Since 3 = 12/4, we have: 12n/4 = 3n/4 + 9 Subtract 3n/4 from each side: 9n/4 = 9 [URL='https://www.mathcelebrity.com/prop.php?num1=9n&num2=9&den1=4&den2=1&propsign=%3D&pl=Calculate+missing+proportion+value']Typing this equation into the search engine[/URL], we get: [B]n = 4[/B]

what integer is used for losing 10 yards A loss means a negative, so a 10 yard loss is written as [B]-10[/B]

What pair of consecutive integers gives the following: 7 times the smaller is less than 6 times the larger? Let x and y be consecutive integers, where y = x + 1 We have 7x < 6y as our inequality. Substituting x, y = x + 1, we have: 7x < 6(x + 1) 7x < 6x + 6 Subtracting x from each side, we have: x < 6, so y = 6 + 1 = 7 (x, y) = (6, 7)

When 3 consecutive positive integers are multiplied, the product is 16 times the sum of the 3 integers. What is the difference of the product minus the sum? Let the 3 consecutive positive integers be: [LIST=1] [*]x [*]x + 1 [*]x + 2 [/LIST] The product is: x(x + 1)(x + 2) The sum is: x + x + 1 + x + 2 = 3x + 3 We're told the product is equivalent to: x(x + 1)(x + 2) = 16(3x + 3) x(x + 1)(x + 2) = 16 * 3(x + 1) Divide each side by (x + 1) x(x + 2) = 48 x^2 + 2x = 48 x^2 + 2x - 48 = 0 Now subtract the sum from the product: x^2 + 2x - 48 - (3x + 3) [B]x^2 - x - 51[/B]

You and a friend want to start a business and design t-shirts. You decide to sell your shirts for $15 each and you paid $6.50 a piece plus a $50 set-up fee and $25 for shipping. How many shirts do you have to sell to break even? Round to the nearest whole number. [U]Step 1: Calculate Your Cost Function C(s) where s is the number of t-shirts[/U] C(s) = Cost per Shirt * (s) Shirts + Set-up Fee + Shipping C(s) = $6.50s + $50 + $25 C(s) = $6.50s + 75 [U]Step 2: Calculate Your Revenue Function R(s) where s is the number of t-shirts[/U] R(s) = Price Per Shirt * (s) Shirts R(s) = $15s [U]Step 3: Calculate Break-Even Point[/U] Break Even is where Cost = Revenue. Set C(s) = R(s) $6.50s + 75 = $15s [U]Step 4: Subtract 6.5s from each side[/U] 8.50s = 75 [U]Step 5: Solve for s[/U] [URL='https://www.mathcelebrity.com/1unk.php?num=8.50s%3D75&pl=Solve']Run this through our equation calculator[/URL] to get s = 8.824. We round up to the next integer to get [B]s = 9[/B]. [B][URL='https://www.facebook.com/MathCelebrity/videos/10156751976078291/']FB Live Session[/URL][/B]

You have $80. Jeans cost $29 and shirts cost $12. Mom told you to buy one pair of jeans and use the rest of the money to buy shirts. Find the inequality. Let j be the number of jeans. Let s be the number of shirts. We are given: [LIST] [*]Mom told you to buy one pair of jeans. So we have $80 to start with - $29 for 1 pair of jeans = $51 left over [/LIST] Now, since shirts cost $12 each, and our total number of shirts we can buy is s, our inequality is [B]12s <= 51[/B]. We want to find the s that makes this inequality true. [URL='https://www.mathcelebrity.com/interval-notation-calculator.php?num=12s%3C%3D51&pl=Show+Interval+Notation']Run this statement through our calculator[/URL], and we get s <= 4.25. But, we need s to be an integer, so we have s <= 4.

You have to pay 29 a month until you reach 850 how many months will that take. Let m be the number of months. We set up the inequality: 29m > = 850 <-- We want to know when we meet or exceed 850, so we use greater than or equal to [URL='https://www.mathcelebrity.com/interval-notation-calculator.php?num=29m%3E%3D850&pl=Show+Interval+Notation']Type this inequality into our search engine[/URL], and we get: m >= 29.31 We round up to the next integer month, to get [B]m = 30[/B].

{x | x is an even integer between -3 and 5} We list even integers out in this range: [B]{-2, 0, 2, 4}[/B]