You need to write down the full question
Megan:
x to the one third power =

<span>x to the one twelfth power = </span>

<span>The quantity of x to the one third power, over x to the one twelfth power is:
</span>

<span>
Since </span>

then

Now, just subtract exponents:
1/3 - 1/12 = 4/12 - 1/12 = 3/12 = 1/4

Julie:
x times x to the second times x to the fifth = x * x² * x⁵
<span>The thirty second root of the quantity of x times x to the second times x to the fifth is
</span>
![\sqrt[32]{x* x^{2} * x^{5} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B32%5D%7Bx%2A%20x%5E%7B2%7D%20%2A%20x%5E%7B5%7D%20%7D%20)
<span>
Since </span>

Then
![\sqrt[32]{x* x^{2} * x^{5} }= \sqrt[32]{ x^{1+2+5} } =\sqrt[32]{ x^{8} }](https://tex.z-dn.net/?f=%5Csqrt%5B32%5D%7Bx%2A%20x%5E%7B2%7D%20%2A%20x%5E%7B5%7D%20%7D%3D%20%5Csqrt%5B32%5D%7B%20x%5E%7B1%2B2%2B5%7D%20%7D%20%3D%5Csqrt%5B32%5D%7B%20x%5E%7B8%7D%20%7D)
Since
![\sqrt[n]{x^{m}} = x^{m/n} }](https://tex.z-dn.net/?f=%20%5Csqrt%5Bn%5D%7Bx%5E%7Bm%7D%7D%20%3D%20x%5E%7Bm%2Fn%7D%20%7D%20)
Then
![\sqrt[32]{ x^{8} }= x^{8/32} = x^{1/4}](https://tex.z-dn.net/?f=%5Csqrt%5B32%5D%7B%20x%5E%7B8%7D%20%7D%3D%20x%5E%7B8%2F32%7D%20%3D%20x%5E%7B1%2F4%7D%20)
Since both Megan and Julie got the same result, it can be concluded that their expressions are equivalent.
Answer:
Step-by-step explanation:
The answer is (B), because the rate of the function is 0.50 from 0 to 8 ounces, thus the line should have a slope of 1/2. But afterward, it levels off horizontally as every ounce above 8 ounces to 12 ounces is $4.
Hope that helps!
Just google it its way easy
Answer:
A=4, B=3, C=2, D=7 or A=2, B=7, C=4, D=3
Step-by-step explanation:
To factor this, you are looking for factors of 8×21 = 168 that have a sum of 34.
168 = 1×168 = 2×84 = 3×56 = 4×42 = 6×28 = 7×24 = 8×21 = 12×14
The sums of these factor pairs are 169, 86, 59, 46, 34, 31, 29, 26. The factor pair whose sum is 34 is 6×28. This means we can rewrite the expression as ...
8x² +28x +6x +21
Factoring by pairs gives ...
4x(2x +7) +3(2x +7)
(4x +3)(2x +7) ⇒ A=4, B=3, C=2, D=7 or A=2, B=7, C=4, D=3