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Hatshy [7]
3 years ago
7

How hard is the permit test?

Mathematics
1 answer:
saveliy_v [14]3 years ago
5 0
It shouldn't be too hard, I've been taking my practice test all day today and yesterday night and I believe i'm going to do good today on my test, just study hard and take the practice test
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A farmer had a farm with animals.
Gnom [1K]

Answer: Part B

Step-by-step explanation:

3 0
3 years ago
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27/64A bottled water company has designed a new cup for its dispenser. The cup will be a right circular cone with a three-inch r
tankabanditka [31]

Answer:

9.86 inches

Step-by-step explanation:

Given:

The cup will be a right circular cone with:-

Radius, r = 3 inches

Volume of cone = 93 cubic inches

Height of the cup, h = ?

Solution:

<u>By using :-</u>

<u />Volume\ of\ right\ circular\ cone=\frac{1}{3} \pi r^{2} h

                                            93 =\frac{1}{3} \times\frac{22}{7} \times3\times3\times h\\ \\ 93=\frac{198}{21} h\\ \\ 93=9.43h

By dividing both sides by 9.43

\frac{93}{9.43} =\frac{9.43h}{9.43} \\ \\ 9.86\ inches=h

Thus, the cup need to be 9.86 inches taller to hold 93 cubic inches of water.

6 0
3 years ago
1. If we have 3 babies what is the probability they’re going to be all boys?
adelina 88 [10]
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3 years ago
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3 years ago
Find the indefinite integral using the substitution provided.
Nady [450]

Answer:  7\text{Ln}\left(e^{2x}+10\right)+C

This is the same as writing 7*Ln( e^(2x) + 10) + C

=======================================================

Explanation:

Start with the equation u = e^{2x}+10

Apply the derivative and multiply both sides by 7 like so

u = e^{2x}+10\\\\\frac{du}{dx} = 2e^{2x}\\\\7\frac{du}{dx} = 7*2e^{2x}\\\\7\frac{du}{dx} = 14e^{2x}\\\\7du = 14e^{2x}dx\\\\

The "multiply both sides by 7" operation was done to turn the 2e^(2x) into 14e^(2x)

This way we can do the following substitutions:

\displaystyle \int \frac{14e^{2x}}{e^{2x}+10}dx\\\\\\\displaystyle \int \frac{1}{e^{2x}+10}14e^{2x}dx\\\\\\\displaystyle \int \frac{1}{u}7du\\\\\\\displaystyle 7\int \frac{1}{u}du\\\\\\

Integrating leads to

\displaystyle 7\int \frac{1}{u}du\\\\\\7\text{Ln}\left(u\right)+C\\\\\\7\text{Ln}\left(e^{2x}+10\right)+C\\\\\\

Be sure to replace 'u' with e^(2x)+10 since it's likely your teacher wants a function in terms of x. Also, do not forget to have the plus C at the end. This is a common mistake many students forget to do.

To verify the answer, you can apply the derivative to it and you should get back to the original integrand of \frac{14e^{2x}}{e^{2x}+10}

4 0
2 years ago
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