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3 years ago

10

Pls answer all questions

2 answers:

Lana71 [14]3 years ago

7 0

15
46
57
13
16
19
71
I hope I can help lol

blagie [28]3 years ago

4 0

1. x=14
2. x=5
3. x=2.5
4.x=9
5.x=5
6.x=13
7.x=-2
8.x=-2
9.x=-10
10.x=13
11.x=-3
12.x=-11
13.x=0
14.x=0.58
15.x=-1.6

hope this helps :)

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Write the equation of a line that is perpendicular to the given line and that passes through the given point. y=3/4x-9;(-8,-18)

d1i1m1o1n [39]

Answer:

<h2>please make me brainlieast and follow me.....</h2>

8 0

3 years ago

Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t

Anika [276]

Answer:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

Step-by-step explanation:

Given

$\int\limits {\frac{x}{x^4 + 16}} \, dx$

Required

Solve

Let

$u = \frac{x^2}{4}$

Differentiate

$du = 2 * \frac{x^{2-1}}{4}\ dx$

$du = 2 * \frac{x}{4}\ dx$

$du = \frac{x}{2}\ dx$

Make dx the subject

$dx = \frac{2}{x}\ du$

The given integral becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

Recall that: $u = \frac{x^2}{4}$

Make $x^2$ the subject

$x^2= 4u$

Square both sides

$x^4= (4u)^2$

$x^4= 16u^2$

Substitute $16u^2$ for $x^4$ in $\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du$

Simplify

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{2}{16}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

In standard integration

$\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)$

So, the expression becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)$

Recall that: $u = \frac{x^2}{4}$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

4 0

3 years ago

Please view the attachment.

quester [9]

-1 * (-4) = 4; 3 * (-4) = -12; 6 * (-4) = -24; -3 * (-4) = 12

7 0

3 years ago

15 POINTSSS AND BRAINLIEST

Ludmilka [50]

1.) 52:97.5
2.) 1:1.875

Hope this helps!

5 0

3 years ago

Need help asap!!! (will give BRAINLEIST)

madam [21]

Answer:

false

Step-by-step explanation:

rainbow method or tabular method x^2+4x+4

7 0

3 years ago

Read 2 more answers