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tino4ka555 [31]
3 years ago
9

HELP ASAP

Mathematics
2 answers:
alexandr402 [8]3 years ago
6 0

Answer:

0.2h

Step-by-step explanation:

13/78 = 0.16666666666667

Rounded to the nearest tenth is the 0.2

Bad White [126]3 years ago
4 0

The answer is 6 and heres how:

78/13=6

hope this helps

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Will give brainliest :,)
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How much would the investment be worth?

As the function for interest is already given to us, also,

The principal amount, P = $7,000

The rate of Interest, r = 3%

Time period, t = 5 years

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Substitute the values,

Hence, the worth of the investment after 5 years at an interest of 3% is $8,123.79.

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Find the explicit formula for the given sequence 2, 4, 8, 16...
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Sarah read 64 1/2 pages of her book in 6 hours. What is the average number of pages that Sarah reads per hour?
MAVERICK [17]

Answer:

We conclude that the average number of pages that Sarah reads per hour will be: 10.75 page per hour

Step-by-step explanation:

Given

  • Number of pages Sarah read =64\frac{1}{2}=\frac{129}{2} = 64.5
  • Number of hours Sarah spent = 6 hours

The average number of pages per hour can be calculated by simply dividing the total number of pages i.e. 64.5 pages by the number of hours spent on reading i.e. 6 hours.

Thus,

Average number of pages per hour = 64.5 pages / 6 hours

                                                            = 10.75 page per hour

Therefore, we conclude that the average number of pages that Sarah reads per hour will be: 10.75 page per hour

8 0
3 years ago
What is the antiderivative of e^x^2?
ddd [48]

Answer: The antiderivative of e^{x^{2}} is \sqrt{\pi} \sqrt{e^{x^2}-1} +C.

Explanation:

I=\int{e^{x^2}dx}\\I^2=\int{e^{x^2}dx}\int{e^{x^2}dx}

Using dawson integral.

I^2=\int{e^{x^2}dx}\int{e^{y^2}dy}

I^2=\int{\int{e^{x^2+y^2}dxdy}}

Put x^2+y^2=r^2

I^2=\int{\int{e^{r^2}rdrd\theta}}

\int_{0}^{2\pi}{d\theta}\int_{0}^{x}{re^{r^2}dr}

Use substitution method and sunbstitute r^2=t.

\int_{0}^{2\pi}{d\theta}\int_{0}^{x^2}{\frac{1}{2}e^tdt}

I^2=\frac{1}{2}|\theta|_{0}^{2\pi}|e^t|_{0}^{x^2}\\I^2=\frac{1}{2}(2\pi)(e^{x^2}-1)\\I=\sqrt{\pi}\sqrt{e^{x^2}-1}.

Therefore, the antiderivative of e^{x^{2}} is \sqrt{\pi} \sqrt{e^{x^2}-1} +C.


4 0
4 years ago
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