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3 years ago

Determine the derivative of y=x^2+xy

Mathematics

1 answer:

ratelena [41]3 years ago

7 0

Hey I just started this test to that's funny lol

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If f(x)=|3x-4|+2 find f(-10)

Nat2105 [25]

Answer:

$f(-10)=36$

Step-by-step explanation:

$f(x)=|3x-4|+2\\f(-10)=|3(-10)-4|+2\\f(-10)=|-30-4|+2\\f(-10)=|-34|+2\\f(-10)=34+2\\f(-10)=36$

Rgds!

6 0

3 years ago

The factory workers make 756 bows in 36 hours. Suppose the workers make the same number of bows each hour. How many bows do they

iren2701 [21]

Answer:

21 bows per hour

Step-by-step explanation:

when your trying to find the perfect hour you divide.

so 756÷36=21

hope this helps

7 0

3 years ago

Read 2 more answers

Circle R has equation

Rudiy27

Answer:

The center is (-10,10) and the radius is 4sqrt(3)

Step-by-step explanation:

(x + 10)^2 + (y - 10)^2 = 48

We can write the equation of a circle as

(x -h)^2 + (y - k)^2 = r^2 where (h,k) is the center and r is the radius

(x- -10)^2 + (y - 10)^2 = (sqrt(16*3) )^2

(x- -10)^2 + (y - 10)^2 = (4sqrt(3)) ^2

The center is (-10,10) and the radius is 4sqrt(3)

5 0

3 years ago

Someone help with this please

r-ruslan [8.4K]

The Answer is B. I hope this helped.

6 0

3 years ago

Find derivative problem<br> Find B’(6)

dalvyx [7]

Answer:

$B^\prime(6) \approx -28.17$

Step-by-step explanation:

We have:

$\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)$

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

$\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]$

We can move the constant outside:

$\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]$

Now, we will utilize the product rule. The product rule is:

$(uv)^\prime=u^\prime v+u v^\prime$

We will let:

$\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t$

Then:

$\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1$

(The derivative of u was determined using the chain rule.)

Then it follows that:

$\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}$

Therefore:

$\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]$

By simplification:

$\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17$

So, the slope of the tangent line to the point (6, B(6)) is -28.17.

5 0

3 years ago