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Molodets [167]
3 years ago
6

Which equation shows the commutative property of addition?

Mathematics
1 answer:
kipiarov [429]3 years ago
7 0

Answer:

C

Step-by-step explanation:

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What is the cube root of 64x12?<br><br> 4x4<br> 4x6<br> 8x4<br> 8x6
vodomira [7]
As 64 is a perfect cube, the coefficient part does not present any problems. However, for the variable with an exponent, it is easier to obtain the cube root by multiplying the exponent by (1/3), since the cube root operator is equivalent to raising a number to the power of (1/3). This is shown below:

cube root (64x^12) = 4 * x^(12)(1/3) = 4 *x^4

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3 years ago
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2x + 5 = 8x - 13 Solve the equation. Do not use decimals in your awnser
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Answer:

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Step-by-step explanation:

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Factor 2a^2-28ab+98b^2
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3 years ago
Which operation results in an expression equivalent to ? 23y^4 - 6y^3 + 35y^2 - 20y
Tom [10]

Answer:

D. P + Q

Explanation:

Given:

P = 8y⁴ + 6y³ + 8y

Q = (5y² - 4y)(3y² + 7)

Required:

Determine which operation will give an expression equivalent to 23y⁴ - 6y³ + 35y² - 20y.

SOLUTION:

Perform each operation given to see which of them gives an expression equivalent to 23y⁴ - 6y³ + 35y² - 20y.

Q - P:

P = 8y⁴ + 6y³ + 8y

Q = (5y² - 4y)(3y² + 7)

Q - P = (5y² - 4y)(3y² + 7) - (8y⁴ + 6y³ + 8y)

(5y²(3y² + 7) -4y(3y² + 7)) - (8y⁴ + 6y³ + 8y)

(15y⁴ + 35y² - 12y³ - 28y) - (8y⁴ + 6y³ + 8y)

Open parentheses

15y⁴ + 35y² - 12y³ - 28y - 8y⁴ - 6y³ - 8y

Collect like terms

15y⁴ - 8y⁴ - 12y³ - 6y³ + 35y² - 28y - 8y

7y⁴ - 18y³ + 35y² - 36y

Therefore, P - Q does not give us an expression equivalent to 23y⁴ - 6y³ + 35y² - 20y.

PQ:

P = 8y⁴ + 6y³ + 8y

Q = (5y² - 4y)(3y² + 7) = (15y⁴ + 35y² - 12y³ - 28y)

PQ = (8y⁴ + 6y³ + 8y)(15y⁴ + 35y² - 12y³ - 28y)

=(8y⁴(15y⁴ + 35y² - 12y³ - 28y) +6y³(8y⁴ + 6y³ + 8y)(15y⁴ + 35y² - 12y³ - 28y) +8y(8y⁴ + 6y³ + 8y)(15y⁴ + 35y² - 12y³ - 28y))

= 120y⁸ . . . . .

Note: You don't need to perform this operation further anymore. It would definitely not give us the equivalent expression we are looking for. The degree of the leading term (120y⁸) is way too greater than the degree of the leading term (23y⁴) of the equivalent expression we are looking for.

Thus, PQ cannot give us an expression equivalent to 23y⁴ - 6y³ + 35y² - 20y.

P - Q:

P = 8y⁴ + 6y³ + 8y

Q = (5y² - 4y)(3y² + 7) = (15y⁴ + 35y² - 12y³ - 28y)

P - Q = (8y⁴ + 6y³ + 8y) - (15y⁴ + 35y² - 12y³ - 28y)

Open parentheses

8y⁴ + 6y³ + 8y - 15y⁴ - 35y² + 12y³ + 28y

Collect like terms

8y⁴ - 15y⁴ + 6y³ + 12y³ - 35y² + 8y + 28y

-7y⁴ + 18y³ - 35y² + 36y

Therefore, P - Q cannot give us an expression equivalent to 23y⁴ - 6y³ + 35y² - 20y.

P + Q:

P = 8y⁴ + 6y³ + 8y

Q = (5y² - 4y)(3y² + 7) = (15y⁴ + 35y² - 12y³ - 28y)

P + Q = (8y⁴ + 6y³ + 8y) + (15y⁴ + 35y² - 12y³ - 28y)

Open parentheses

8y⁴ + 6y³ + 8y + 15y⁴ + 35y² - 12y³ - 28y

Collect like terms

8y⁴ + 15y⁴ + 6y³ - 12y³ + 35y² + 8y - 28y

23y³ - 6y³ + 35y² - 20y

Therefore, P + Q will result in an expression equivalent to 23y⁴ - 6y³ + 35y² - 20y.

The answer is D.

7 0
3 years ago
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