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Citrus2011 [14]

3 years ago

12

Write an equation in point-slope form of the line having the given slope that contains the given point. m = -2, (4, -2)write an

equation in point slope form of the line having the given slope that contains the given point m equals negative 2, (4,-2)

1 answer:

alex41 [277]3 years ago

6 0

Answer:

Point-slope form: y+2=-2(x-4)

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Solve for y<br> (y-k)h=c

Shalnov [3]

Answer:

y=(x+kh)1/h

Step-by-step explanation:

(y-k)h=c

hy-kh=c

hy=x+kh

y=(x+kh)1/h

4 0

3 years ago

Evaluate 0.3 + y/2 when y = 10 and z =5

Vladimir79 [104]

Answer:

0.3*10+10/5

PEMDAS (left to right)

3+2

5

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

The inverse, f'(x), of f(x) =

astraxan [27]

Step-by-step explanation:

Given

$f(x) = \frac{x + 5}{x}$

So

$f(2) = \frac{2 + 5}{2} \\ = \frac{7}{2}$

Now

${f}^{ - 1} (x) = \frac{5}{x - 1}$

So

${f}^{ - 1} (2) = \frac{5}{2 - 1} \\ = \frac{5}{ 1} \\ = 5$

Hope it will help :)❤

7 0

3 years ago

Peter uses the equation y = StartFraction 13 over 4 EndFraction x to model the number of miles that he has walked in x hours. Wh

Umnica [9.8K]

The true statement is Peter walks at a rate of 13 over 4 miles per hour.

<h3>What is the true statement?
</h3>

Direct variation is when two variables move in the same direction. If one variable increases, the other variable increases. When the hour Peter walks increases, the distance he walks also increases.

Here are the options:

Peter walks at a rate of StartFraction 4 over 13 EndFraction miles per hour.

Peter walks at a rate of 4 miles per hour.

Peter walks at a rate of StartFraction 13 over 4 EndFraction miles per hour.

Peter walks at a rate of 13 miles per hour.

To learn more about direct variation, please check: brainly.com/question/27573249

#SPJ1

8 0

2 years ago

Find the general indefinite integral. (use c for the constant of integration.) (7θ − 6 csc(θ) cot(θ)) dθ

KiRa [710]

$\int \! 70 - 6 \csc(\theta) \cot(\theta) \ \mathrm{d}\theta = \int \! 70 - 6 \cdot \frac{1}{\sin(\theta)} \cdot \frac{\cos(\theta)}{\sin(\theta)} \ \mathrm{d}\theta = \int \! 70 \ \mathrm{d}\theta - \int \! \frac{6 \cos(\theta)}{\sin^2(\theta)} \ \mathrm{d}\theta = 70 \theta + \frac{6}{\sin(\theta)} + C, \ C \in \mathbb{R}.$

8 0

4 years ago