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trapecia [35]
4 years ago
11

for the normal (perpendicular) line to the curve y= square root of 8-x^2 at (-2,2) would the slope be 1/2? ...?

Mathematics
1 answer:
Umnica [9.8K]4 years ago
6 0
The correct answer to this question is -1.
The equation is 
y= sqrt(8-x^2) 
y' = -x/sqrt(8-x^2) 
y'(-2) = 2/2 = 1 
the normal would have the slope -1 f<span>or the normal (perpendicular) line to the curve y= square root of 8-x^2 at (-2,2). 

Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help. 
</span>
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Which of the following can not be used to calculate 32% of 50
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Answer:

the answer is 16

Step-by-step explanation:

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3 years ago
You are jumping off the 12 foot diving board at the municipal pool. You bounce up at 6 feet per second and drop to the water you
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Answer:

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1.075 seconds after you jump.

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Step-by-step explanation:

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h(t) = 0

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Here we just need to find the value of t such that:

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t = \frac{-6 \pm \sqrt{6^2 - 4*(-16)*12} }{2*(-16)} = \frac{-6 \pm 28.4}{-32}

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