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2 years ago

5

Someone do this for me plz

2 answers:

kumpel [21]2 years ago

4 0

Iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiioi

sladkih [1.3K]2 years ago

3 0

Answer:

(all in standard form)

1. $6x^2-x+4$

2. $-3x^2-4x+6$

3. $s^3+s^2+4s+1$

4. $5a^3-a^2-2a+10$

5. $a^4+a^3-3a^2+4$

6. $u^3-u^2+9u+2$

7. $-8u^2+5u-2$

8. $81y^3+3y^2+9y+18$

9. $-3a^2+5a-2$

10. $21^3+2b^2+4b-8$

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1/3 + 1/9 need help fast help

nikdorinn [45]

Answer:

the answer is 4/9

Step-by-step explanation:

1/3= 3/9

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2 years ago

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Simplify this expression<br><br> (3 + i) - (2- 2i)

ivolga24 [154]

1+3i

like standart algebra

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2 years ago

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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

2 years ago

12. What is the slope of the line <br> =3x + 5?<br><br> 5<br> 3<br> -5<br> -3

Reika [66]

Answer:

3, 3x+5

Step-by-step explanation:

DIFFERENTIATE W.R.T. X

3

EVALUATE

3x+5

4 0

2 years ago

I need help plss. I will give you a lot of points I promise

Alekssandra [29.7K]

Slope=0.005/2.000=0.002
p - intercept = 974/1 = 974.00000
n- intercept = 974/-404 = 487/-202 = -2.41089

Oh and it is ok if you don’t have a lot of points I am great full for the points you are giving me and the answer is C)13,094 and can I have the brainliest since the other guy clearly did not answer the question right I only have one brainliest ☺️

8 0

2 years ago

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