(2a^(3))^(-3)
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(3b(-2))

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Answer 1

Answer: Hello,

$((2a^(3))^(-3))/(3b*(-2))=-(1)/((2a^3)^3*6b)=-(1)/(48a^9b)$

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Order the set of integers from greatest to least. {-100, -89, -124, -69, -52}

Answers

Answer:

-52, -69, -89, -100, -124

Step-by-step explanation:

When your dealing with negative numbers the smaller number the larger it is.

Hope this helps have a nice day:)

Use the difference of cubes for 27x^(3)-1

Answers

Note:

$(a-b)^3=a^3-3a^2b+3ab^2-b^3\n\mathrm{or,\ }(a-b)^3=a^3-b^3-3ab(a-b)\n\mathrm{or,\ }a^3-b^3=(a-b)^3+3ab(a-b)\n\mathrm{or,\ }a^3-b^3=(a-b)[(a-b)^2+3ab]\n\mathrm{or,\ }a^3-b^3=(a-b)(a^2-2ab+b^2+3ab)\n\therefore\ a^3-b^3=(a-b)(a^2+ab+b^2)$

Answer:

$27x^3-1\n=(3x)^2-1\n=(3x-1)[(3x)^2+(3x)(1)+1^2]\n=(3x-1)(9x^2+3x+1)$

Answer:

To factor the expression 27x^3 - 1 using the difference of cubes formula, we can follow these steps:

1. Identify the cube root of each term. In this case, the cube root of 27x^3 is 3x, and the cube root of 1 is 1.

2. Write the formula for the difference of cubes, which is: a^3 - b^3 = (a - b)(a^2 + ab + b^2).

3. Replace "a" with 3x and "b" with 1 in the formula.

(3x)^3 - 1^3 = (3x - 1)((3x)^2 + (3x)(1) + 1^2)

4. Simplify the expression inside the parentheses.

(3x - 1)(9x^2 + 3x + 1)

Therefore, using the difference of cubes formula, we can factor 27x^3 - 1 as (3x - 1)(9x^2 + 3x + 1).

Step-by-step explanation:

How do you " find, to the nearest tenth of an inch, the diagonal of a square whose perimeter is 28 inches ?"

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Well if the perimeter is 28 inches, each side of the square is 7 inches. 28/4 = 7 because all sides of a square are the same.

If we split the square in half to create 2 triangles we can find the length of the diagonal. The two sides are the bases of 7 inches.

Use the Pythagorean theorem to solve for the hypotenuse or the diagonal.

7^2 + 7^2 = diagonal ^ 2

Therefore, the diagonal equals 9.9 inches.

Which represents the solution(s) of the system of equations, y = –x2 + 6x + 16 and y = –4x + 37? Determine the solution set algebraically.a.(3, 25)
b.(–3, 49)
c.(3, 25) and (7, 9)
d.(–3, 49) and (–7, 65)

Answers

Answer: c.(3, 25) and (7, 9)

y = –x^2 + 6x + 16 and y = –4x + 37

Plug in -4x+37 for y in first equation . It becomes

$-x^2 + 6x + 16= -4x+37$

Combine like terms. add 4x and subtract 37 on both sides

$-x^2 + 10x - 21=0$

Divide the whole equation by -1 to remove negative sign from -x^2

$x^2 - 10x + 21=0$

Now factor the left hand side

(x-7)(x-3) = 0

x-7 =0 and x-3=0

x= 7 and x=3

Now we find out y using y = –4x + 37

when x= 7 , then y=-4(7) +37 = 9

when x= 3, then y=-4(3) + 37 = 25

We write solution set as (x,y)

(7,9) and (3,25) is our solution set

TO determine the solution set of the equations given above, we can use substitution method. We do as follows:

y = –x2 + 6x + 16 and y = –4x + 37

–x2 + 6x + 16 = –4x + 37
-x^2 +10x -21 = 0
x = 7 and 3
y = 9 and 25

Therefore, option C is the correct answer.

Solve for x
Ax + B = C

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Answer:

(C - B)/A = x

Step-by-step explanation:

Ax + B = C

-B -B

Ax = C - B

Divide both sides by A

x = (C - B)/A

(C - B)/A = x

Find the missing side and round to the nearest tenth

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In this problem you use cosine because you know the hypotenuse and you want to know the adjacent side of the triangle. So in your calculator you would input cos(52). Then you would multiply that answer with the hypotenuse side. So your equation would be this: cos(52) x 13

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(2a^(3))^(-3) -------- (3b(-2))

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(2a^(3))^(-3)
--------
(3b(-2))