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vivado [14]
3 years ago
15

Find the derivative of y=8ln(sin t+9)

Mathematics
1 answer:
ira [324]3 years ago
6 0
\bf y=8ln[sin(t+9)]\\\\
-----------------------------\\\\
\cfrac{dy}{dx}=8\cdot \cfrac{cos(t+9)\cdot 1}{sin(t+9)}\implies \cfrac{dy}{dx}=8cot(t+9)
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FIRST ANSWER GETS BRANLIEST!!
Mashcka [7]

Answer:

(0,0)

Step-by-step explanation:

Let's think about how the graph of this would look.

We know that y will be x^2, meaning that if x was 0, y would be 0.

If x was 1, y would be 1

If x was 2, y would be 4.

Now let's think about negative numbers.

If x was -1, y would be 1 (since -1^2 = 1

If x was -2, y would be 4 (since -2^2 = 4).

This means that our graph would turn at the point where x and y are equal (excluding 1,1)

This is 0,0.

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Hope this helped!

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answer two numbers are 6 and 9
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