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3 years ago

Given g(x)=5x+2, find g(-6)

2 answers:

7 0

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Temka [501]3 years ago

5 0

Answer:

Please refer to the attachment

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A car is reduced by 20% in price to $40,000. What was the original price?

VikaD [51]

48000 i think i might have missed a step

8 0

3 years ago

Read 2 more answers

Rewrite expression as a single power<img src="https://tex.z-dn.net/?f=6%5E%7B4%7D%2A%206%5E%7B4%7D" id="TexFormula1" title="6^{4

victus00 [196]

Answer:

6^8

Step-by-step explanation:

6^4 * 6^4

We know that a^b * a^c = a^( b+c)

6^(4+4)

6^8

5 0

3 years ago

Read 2 more answers

HELP!!!

vladimir1956 [14]

Create a conversion factor from "There are 6.02 × 10^23 atoms in 1 gram of hydrogen:"

6.02*10^23 atoms
-------------------------
1 gram

Then mult. 800 grams by this conversion factor:

6.02*10^23 atoms 800 grams
------------------------- * ----------------
1 gram 1

This comes out to 4816*10^23 atoms, which, in scientific notation, is

4.816*10^26 atoms.

3 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

Use the markings in the image to answer the following questions (drawing not to scale):

olga nikolaevna [1]

Hope this helps!

There are 3 images with full explanation

3 0

3 years ago