Answer:
f'(x) = -f(x) = 9xy² - 6x²y + 5x³
Step-by-step explanation:
f(x) = –9xy² + 6x²y – 5x³
additive inverse: f'(x) = -f(x) = 9xy² - 6x²y + 5x³
f(x) + f'(x) = f(x) -f(x) = (–9xy² + 6x²y – 5x³) + (9xy² - 6x²y + 5x³) = 0
The answer to the question is B
Answer:
45.1feet
Step-by-step explanation:
Given the following
∠I=90°
∠G=62°, and
GH = 96 feet = Hypotenuse
Required
IG = Adjacent side
Using the SOH CAH TOA identity
Cos theta = Adj/hyp
Cos 62 =IG/96
IG = 96cos62
IG = 96(0.4695)
IG = 45.1feet
Hence the length of IG to the nearest tenth is 45.1feet
Answer:
B
Step-by-step explanation:
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Step-by-step explanation:
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