Question

Help me with these math questions......

Answer 1

Answer: y = 0

Explanation:

`(x^(0))/(x^(2))`

exponent of denominator > exponent of numerator so, y = 0

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Answer:
`y=(x^(2)+8)/(3)`

Explanation:

y =
`√(3x-8)`

To find the inverse, swap the x's and y's and solve for "y":

x =
`√(3y-8)`

x² = 3y - 8

x² + 8 = 3y

`(x^(2)+8)/(3)=y`

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Answer: -6x - 3h

Explanation:

f(x + h) = 7 - 3(x + h)²

= 7 - 3(x² + 2xh + h²)

= 7 - 3x² - 6xh - 3h²

-f(x) = -(7 - 3x²)

= -7 + 3x²

f(x + h) - f(x) = 7 - 3x² - 6xh - 3h² -7 + 3x²

= -6xh - 3h²

= h(-6x - 3h)

`(f(x+h)-f(x))/(h)=(h(-6x-3h))/(h)`

= -6x - 3h