(n^2)^3 without exponents(n^2)^3 without exponents
This expression evaluates to:
n^(2 *3)
n^6
To write this without exponents, we expand n times itself 6 times:
[B]n * n * n * n * n * n
[MEDIA=youtube]zVAlzX9oHOQ[/MEDIA][/B]
3abc^4/12a^3(b^3c^2)^2 * 8ab^-4c/4a^2b3abc^4/12a^3(b^3c^2)^2 * 8ab^-4c/4a^2b
Expand term 1:
3abc^4/12a^3(b^3c^2)^2
3abc^4/12a^3b^6c^4
Now simplify term 1:
3/12 = 1/4
c^4 terms cancel
Subtract powers from variables since the denominator powers are higher:
b^(6 - 1) = b^5
a^(3 - 1) = a^2
1/4a^2b^5
Now simplify term 2:
8ab^-4c/4a^2b
8/4 = 2
2c/a^(2 - 1)b^(1 - -4)
2c/ab^5
Now multiply simplified term 1 times simplified term 2:
1/4a^2b^5 * 2c/ab^5
(1 * 2c)/(4a^2b^5 * ab^5)
2c/4a^(2 + 1)b^(5 + 5)
2c/4a^3b^10
2/4 = 1/2, so we have:
[B]c/2a^3b^10[/B]
A man purchased 20 tickets for a total of $225. The tickets cost $15 for adults and $10 for childrenA man purchased 20 tickets for a total of $225. The tickets cost $15 for adults and $10 for children. What was the cost of each ticket?
Declare variables:
[LIST]
[*]Let a be the number of adult's tickets
[*]Let c be the number of children's tickets
[/LIST]
Cost = Price * Quantity
We're given two equations:
[LIST=1]
[*]a + c = 20
[*]15a + 10c = 225
[/LIST]
Rearrange equation (1) in terms of a:
[LIST=1]
[*]a = 20 - c
[*]15a + 10c = 225
[/LIST]
Now that I have equation (1) in terms of a, we can substitute into equation (2) for a:
15(20 - c) + 10c = 225
Solve for [I]c[/I] in the equation 15(20 - c) + 10c = 225
We first need to simplify the expression removing parentheses
Simplify 15(20 - c): Distribute the 15 to each term in (20-c)
15 * 20 = (15 * 20) = 300
15 * -c = (15 * -1)c = -15c
Our Total expanded term is 300-15c
Our updated term to work with is 300 - 15c + 10c = 225
We first need to simplify the expression removing parentheses
Our updated term to work with is 300 - 15c + 10c = 225
[SIZE=5][B]Step 1: Group the c terms on the left hand side:[/B][/SIZE]
(-15 + 10)c = -5c
[SIZE=5][B]Step 2: Form modified equation[/B][/SIZE]
-5c + 300 = + 225
[SIZE=5][B]Step 3: Group constants:[/B][/SIZE]
We need to group our constants 300 and 225. To do that, we subtract 300 from both sides
-5c + 300 - 300 = 225 - 300
[SIZE=5][B]Step 4: Cancel 300 on the left side:[/B][/SIZE]
-5c = -75
[SIZE=5][B]Step 5: Divide each side of the equation by -5[/B][/SIZE]
-5c/-5 = -75/-5
c = [B]15[/B]
Recall from equation (1) that a = 20 - c. So we substitute c = 15 into this equation to solve for a:
a = 20 - 15
a = [B]5[/B]
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* Binomial Expansions c(a + b)x
* Polynomial Expansions c(d + e + f)x
* FOIL Expansions (a + b)(c + d)
* Multiple Parentheses Multiplications c(a + b)(d + e)(f + g)(h + i)
Expanded NotationFree Expanded Notation Calculator - Writes the expanded notation for a number.
Find two consecutive odd integers such that the sum of their squares is 290Find 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 sum of their squares is 25Find 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
Finding the dimensionsHow do I find dimensions of a rectangle when it has been expanded?
Finding the dimensionsExpanded area = Original Area + area of Expansion
Area of Expansion = length expansion * width expansion
I had a brother but my brother had no brothers. how can this beI had a brother but my brother had no brothers. how can this be
Because "I" is a female.
To solve trick questions like this, you must expand your theory of constraints.
Most people look at this problem and see the word [I]brother [/I]twice and limit themselves to thinking in terms of men.
Last month, a parking lot had 23 spaces in each of its rows. Recently, the lost was expanded, and 4Last month, a parking lot had 23 spaces in each of its rows. Recently, the lost was expanded, and 4 spaces were added to each row. If the lot has 8 rows, how many spaces are there now?
23 spaces + 4 additional spaces = 27 spaces
27 spaces * 8 rows = [B]216 spaces[/B]
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Prove 0! = 1Prove 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
Prove 0! = 1[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
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Sally and Adam works a different job. Sally makes $5 per hour and Adam makes $4 per hour. They eachSally and Adam works a different job. Sally makes $5 per hour and Adam makes $4 per hour. They each earn the same amount per week but Adam works 2 more hours. How many hours a week does Adam work?
[LIST]
[*]Let [I]s[/I] be the number of hours Sally works every week.
[*]Let [I]a[/I] be the number of hours Adam works every week.
[*]We are given: a = s + 2
[/LIST]
Sally's weekly earnings: 5s
Adam's weekly earnings: 4a
Since they both earn the same amount each week, we set Sally's earnings equal to Adam's earnings:
5s = 4a
But remember, we're given a = s + 2, so we substitute this into Adam's earnings:
5s = 4(s + 2)
Multiply through on the right side:
5s = 4s + 8 <-- [URL='https://www.mathcelebrity.com/expand.php?term1=4%28s%2B2%29&pl=Expand']multiplying 4(s + 2)[/URL]
[URL='https://www.mathcelebrity.com/1unk.php?num=5s%3D4s%2B8&pl=Solve']Typing this equation into the search engine[/URL], we get s = 8.
The problem asks for Adam's earnings (a). We plug s = 8 into Adam's weekly hours:
a = s + 2
a = 8 + 2
[B]a = 10[/B]
SequencesFree Sequences Calculator - Given a function a(n) and a count of sequential terms you want to expand (n), this calcuator will determine the first (n) terms of your sequence, {a1, a2, ..., an}
Solve for xExpand the right side:
1/3x + 1/2 = 6/4x - 10
Simplify as 6/4 is 3/2
x/3 + 1/2 = 3x/2 - 10
Common denominator of 2 and 3 is 6. So we have:
2x/6 + 1/2 = 9x/6 -10
Subtract 2x/6 from each side
7x/6 - 10 = 1/2
Add 10 to each side. 10 is 20/2
7x/6 = 21/2
Using our [URL='http://www.mathcelebrity.com/prop.php?num1=7x&num2=21&den1=6&den2=2&propsign=%3D&pl=Calculate+missing+proportion+value']proportion calculator[/URL], we get:
[B]x = 9[/B]
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Standard NotationFree Standard Notation Calculator - Displays the Standard notation of expanded notation
The sum of twice an integer and 3 times the next consecutive integer is 48The 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]
Two numbers have a sum of 20. Determine the lowest possible sum of their squares.Two numbers have a sum of 20. Determine the lowest possible sum of their squares.
If sum of two numbers is 20, let one number be x. Then the other number would be 20 - x.
The sum of their squares is:
x^2+(20 - x)^2
Expand this and we get:
x^2 + 400 - 40x + x^2
Combine like terms:
2x^2 - 40x + 400
Rewrite this:
2(x^2 - 20x + 100 - 100) + 400
2(x - 10)^2 - 200 + 400
2(x?10)^2 + 200
The sum of squares of two numbers is sum of two positive numbers, one of which is a constant of 200.
The other number, 2(x - 10)^2, can change according to the value of x. The least value could be 0, when x=10
Therefore, the minimum value of sum of squares of two numbers is 0 + 200 = 200 when x = 10.
If x = 10, then the other number is 20 - 10 = 10.
Van needs to enter a formula into a spreadsheet to show the outputs of an arithmetic sequence that sVan needs to enter a formula into a spreadsheet to show the outputs of an arithmetic sequence that starts with 13 and continues to add seven to each output. For now, van needs to know what the 15th output will be. Complete the steps needed to determine the 15th term in sequence.
Given a first term a1 of 13 and a change amount of 7, expand the series
The explicit formula for an [I]arithmetic series[/I] is an = a1 + (n - 1)d
d represents the common difference between each term, an - an - 1
Looking at all the terms, we see the common difference is 7, and we have a1 = 13
Therefore, our explicit formula is an = 13 + 7(n - 1)
If n = 15, then we plug it into our explicit formula above:
an = 13 + 7(n - 1)
a(15) = 15 + 7(15 - 1)
a(15) = 15 + 7 * 14
a(15) = 15 + 98
a(15) = [B]113[/B]
Which of the following is equivalent to 3(2x + 1)(4x + 1)?Which of the following is equivalent to 3(2x + 1)(4x + 1)?
[LIST]
[*]A) 45x
[*]B) 24x^2 + 3
[*]C) 24x^2 + 18x + 3
[*]D) 18x^2 + 6
[/LIST]
First, [URL='https://www.mathcelebrity.com/binomult.php?term1=2x%2B1&term2=4x%2B1&pl=Expand+Product+of+2+Binomials+using+FOIL']multiply the binomials[/URL]:
We get 8x^2 + 6x + 1
Now multiply this polynomial by 3:
3(8x^2 + 6x + 1) = [B]24x^2 + 18x + 3, answer C[/B]