V(x) = (2x - 3)/(5x + 4)
The domain is all Real numbers except x = -4/5, because if x = -4/5 the denominator would be zero and you cannot divide by zero.
{x | x ∈ R, x ≠ -4/5}
w(x) = (5x + 4)/(2x - 3)
similarly, x ≠ 3/2
so, {x| x ∈ R, x ≠ 3/2}
Answer:
the answer is d
Step-by-step explanation:
Domain: all real numbers
Range: y<0
Answer:
![\frac{\sqrt[3]{16y^4}}{x^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B16y%5E4%7D%7D%7Bx%5E2%7D)
Step-by-step explanation:
The options are missing; However, I'll simplify the given expression.
Given
![\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B32x%5E3y%5E6%7D%7D%7B%5Csqrt%5B3%5D%7B2x%5E9y%5E2%7D%20%7D)
Required
Write Equivalent Expression
To solve this expression, we'll make use of laws of indices throughout.
From laws of indices ![\sqrt[n]{a} = a^{\frac{1}{n}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%7D%20%20%3D%20a%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D)
So,
gives

Also from laws of indices

So, the above expression can be further simplified to

Multiply the exponents gives

Substitute
for 32


From laws of indices

This law can be applied to the expression above;
becomes

Solve exponents


From laws of indices,
; So,
gives

The expression at the numerator can be combined to give

Lastly, From laws of indices,
; So,
becomes
![\frac{\sqrt[3]{(2y)}^{4}}{x^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B%282y%29%7D%5E%7B4%7D%7D%7Bx%5E2%7D)
![\frac{\sqrt[3]{16y^4}}{x^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B16y%5E4%7D%7D%7Bx%5E2%7D)
Hence,
is equivalent to ![\frac{\sqrt[3]{16y^4}}{x^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B16y%5E4%7D%7D%7Bx%5E2%7D)
Answer:
Addition. “Also, I have to stop at the store on the way home.” ...
Comparison. “In the same way, the author foreshadows a conflict between two minor characters.” ...
Concession. “Granted, you did not ask ahead of time.” ...
Step-by-step explanation: