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Fed [463]
3 years ago
8

The results of a survey indicate 50% of seventh-graders watch at least one movie

Mathematics
2 answers:
Kay [80]3 years ago
7 0

Answer:

25 students

Step-by-step explanation:

50% of 50 is 25  

50/2 = 25

sergey [27]3 years ago
7 0

Answer:25

Step-by-step explanation:

The total number of children is 50

So 50%of50children is

50/100×50=2500/100

=25

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Answer:

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Step-by-step explanation:

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3 years ago
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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
Triangle abc has given the area: 2.95 , angle A: 80 and side b:2. Find the requested length, c
klio [65]
Can you support your question by uploading a drawing?
7 0
3 years ago
Manny is covering a book. The front of the book is 11.2 inches high and 7.3 inches wide. What is the area of the front of the bo
irina [24]

Answer:

81.76in^2

Step-by-step explanation:

Given data

Lenght of book=  11.2 inches

WIdth of book= 7.3 inches

We know that Area= Length* Width

Area= 11.2*7.3

Area= 81.76 in^2

Hence the area of the front of the book is is 81.76in^2

7 0
3 years ago
Why is student A correct?
ASHA 777 [7]

Answer:

area=1/2bh

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150÷10

h=15

4 0
3 years ago
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