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xenn [34]
3 years ago
13

Find the cost of one item to the nearest cent. Round if necessary. a dozen donuts for $4.00

Mathematics
2 answers:
kondaur [170]3 years ago
5 0

Answer:

33 cents

Step-by-step explanation:

4/12=1/3=0.33=33 cents

Molodets [167]3 years ago
4 0

Answer:

= $1/3

Step-by-step explanation:

A dozen contains 12 items

Therefore;

a dozen of doughnuts contains 12 doughnuts

12 doughnuts = $ 4.00

1 doughnuts = 4/12

                    = $ 1/3

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The center of a circle is at (-5 12) and it has a radius of 16. what is the equation of the circle?
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Answer:

(x+5)^2+(y-12)^2=256

Step-by-step explanation:

Comparing the given (x+5)^2+(y-12)^2=256

to                                   (x - h)^2 + (y - k)^2 = r^2,  

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8 0
2 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
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