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3 years ago

If x+9.8=14.7, what is the value of 8(x-3.7)?

Mathematics

2 answers:

Verizon [17]3 years ago

7 0

X=4.9 And 8(4.9-3.7) therefore when we solve it we get answer as 9.6

sergiy2304 [10]3 years ago

3 0

First you need to know what X is. So you need to subtract.
14.7-9.8=4.9 is X
Now you put X in parenthesis.
(4.9-3.7)=1.2 now you multiply 8x 1.2=9.6 is your answer. Always start with PEMDAS and it stands for the order to do the math. Parenthesis, Exponents,
Multiplication,
Division,
Addition and Subtraction
Hope this helps:)

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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:

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