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3 years ago

How do I find one side of an angle

Mathematics

2 answers:

dezoksy [38]3 years ago

8 0

Answer:

lemme just say that they want me to give first answer

Dimas [21]3 years ago

6 0

Answer: Can you explain more? you have to calculate it by an theorm

Step-by-step explanation:

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A newly drilled water well produces 95,000 quarts of water per week. With no new water feeding the well, the production drops by

VARVARA [1.3K]

Step-by-step explanation:

1 week =95000

52 weeks =x

x=95000×52

5 0

3 years ago

Brianna's parents earned $43,329 for the year. During the year they paid $9,966 in taxes. At the end of the year when taxes are

saveliy_v [14]

Answer:

1694

Step-by-step explanation:

The $43,329 in this question is just to trick you, so don't worry about this number.

We know her parents have already payed $9,966 for this years taxes.

But they need to pay a total of $11,660 for the year.

11,660 - 9966

=1694

7 0

3 years ago

Using y=mx+b find the slope for 5) 2x − y =1

frosja888 [35]

Answer:

Step-by-step explanation:

change it to slope intercept form

2x-y=1

add y

2x=y+1

subtract 1

2x-1=y

reorder

y=2x-1

so the slope is 2

7 0

3 years ago

What is equivalent to sq root -27

Katena32 [7]

Negative numbers don’t have real square roots since a square is either positive or 0

7 0

3 years ago

Read 2 more answers

Let f be defined by the function f(x) = 1/(x^2+9)

riadik2000 [5.3K]

(a)

$\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}$

Substitute x = 3 tan(t ) and dx = 3 sec²(t ) dt :

$\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt$

$=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}$

(b) The series

$\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}$

converges by comparison to the convergent p-series,

$\displaystyle\sum_{n=3}^\infty\frac1{n^2}$

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}$

converges absolutely, since

$\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}$

That is, ∑ (-1)ⁿ (n ² + 9)/eⁿ converges absolutely because ∑ |(-1)ⁿ (n ² + 9)/eⁿ| = ∑ (n ² + 9)/eⁿ in turn converges by comparison to a geometric series.

5 0

3 years ago