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3 years ago

Find the slope of the line passing through (-7,-8) and (5,-8)

Mathematics

1 answer:

zhenek [66]3 years ago

8 0

Answer:

slope = 0

Step-by-step explanation:

Calculate the slope m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (- 7, - 8) and (x₂, y₂ ) = (5, - 8)

m = $\frac{-8+8}{5+7}$ = $\frac{0}{12}$ = 0

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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

3 years ago

Help I need answers now

Juli2301 [7.4K]

Answer:

<u>The Answer in Exact Form:</u>

-3/22

<u>In Decimal Form:</u>

-0.136

Step-by-step explanation:

- $\frac{3}{2}$ ÷5 $\frac{1}{2}$ =-3/22

8 0

3 years ago

Monthly water bills for a city have a mean of $108.43 and a standard deviation of $36.98. Find the probability that a randomly s

GaryK [48]

Answer:

The bill size can be considered usual.

Step-by-step explanation:

Mean (μ) = $108.43

Standard deviation (σ) = $36.98

Now, consider the distribution to be normal distribution :

$P(Z=\frac{X-\mu}{\sigma}>\frac{a-\mu}{\sigma})=P(Z>\frac{173-108.43}{36.98})\\\\\implies P(Z>1.75)$

Now, finding values of z-score from the table. We get,

P(Z > 1.75) = 0.9599

⇒ 95.99%

So, only 4.01 % of the people in the city wastes water.

Hence, the bill size can be considered usual.

6 0

3 years ago

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Evaluate 12∣∣−x+3∣∣−5 when x = 11.

Evgen [1.6K]

Answer:

Step-by-step explanation:

12∣∣−x+3∣∣−5

Let x = 11

12∣∣−11+3∣∣−5

Do the absolute value first

12∣∣-8∣∣−5

Taking the absolute value, which means making it positive

12* 8 -5

Multiply and divide

96 -5

7 0

3 years ago

A2 + 7a + 10 = (a + 5)(a + ? )

Brilliant_brown [7]

$a^{2}$ +7a+10 = (a+5)(a+2)

3 0

3 years ago

Read 2 more answers