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Airida [17]
3 years ago
11

Help me please I will give brainiest​

Mathematics
1 answer:
nadya68 [22]3 years ago
8 0
8.a)
A= (1,1) A’= (3,3)
B= (4,6) B’= (12,18)
C= (6,1). C= (18,3)

8b.)

A= (1,1) A’(0,5,0.5)

B= (4,6) B(2,3)

C= (6,1). C(3,0.5)


Hope this helped ! :)
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The total cost (in dollars) for a company to manufacture and sell x items per week is C = 40 x + 640 , whereas the revenue broug
ikadub [295]

Answer:

The company must sell 60 or 70 items to obtain a weekly profit of 200.

Step-by-step explanation:

The profit is the difference between the revenue and the cost of a given task, therefore:

\text{profit} = R - C\\\text{profit} = 66*x - 0.2*x^2 - (40*x + 640)\\\text{profit} = 66*x - 0.2*x^2 - 40*x - 640\\\text{profit} = - 0.2*x^2 + 26*x - 640

To have a profit of 200, we need to sell:

-0.2*x^2 + 26*x - 640 = 200\\-0.2*x^2 + 26*x -840 = 0\text{ } *\frac{-1}{0.2}\\x^2 -130 + -4200 = 0\\x_{1,2} = \frac{-(-130) \pm \sqrt{(-130)^2 - 4*1*(-4200)}}{2*1}\\x_{1,2} = \frac{130 \pm \sqrt{16900 + 16800}}{2}\\x_{1,2} = \frac{130 \pm \sqrt{100}}{2}\\x_{1,2} = \frac{130 \pm 10}{2}\\x_{1} = \frac{130 + 10}{2} = \frac{140}{2} = 70\\ x_{2} = \frac{130 - 10}{2} = \frac{120}{2}  = 60

The company must sell 60 or 70 items to obtain a weekly profit of 200.

3 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.

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