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3 years ago

14

Help I’m being timed and I need someone to do this quickly

2 answers:

S_A_V [24]3 years ago

6 0

Answer: 1

Step-by-step explanation:

It's khan so ik the answer, Trust me

kondor19780726 [428]3 years ago

6 0

Answer:

The answer is 1.

Step-by-step explanation:

If you divide 15 by 3, then the answer would be 5. Since it is a proportional relationship, you have to do the same to the <em>y-axis</em>. Therefore, you must divide the first <em>y-axis </em>by 3 in order to get the second <em>y-axis</em>. Doing this will land you with a second <em>y-axis</em> of one.

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What is the solution to the system of equations?

natka813 [3]

Answer:

Step-by-step explanation:

There is an error in the question. The table does not show two linear functions. y₁ is a linear function, but y₂ is not a straight line. It makes a bend at (-6,1).

Line 1 goes through (-12,-3) and (0,5).

slope = (5-(-3))/(0-(-12)) = 2/3

y-intercept = 5

y₁ = (2/3)x + 5

Line 2 goes through (-12,-2) and (-6,1).

slope = (1-(-2))/(-6-(-12)) = 1/2

y₂ = (1/2)x + 4

(2/3)x + 5 = (1/2)x + 4

x = -6

y = (2/3)x + 5 = 1

Solution: (-6,1)

3 0

3 years ago

Use substitution to solve the following system of equations. What is the value of y?

koban [17]

The answer is C. I solved for the first variable in one of the equations then substitute the result into the other equation.

5 0

3 years ago

A business owner pays $1,200 per month in rent and a total of $120 per hour in employee salary for each hour the store is open.

Cloud [144]

0 = 200x - 120x - 1200

6 0

3 years ago

Help me with differentation and integration please!!

Marina86 [1]

Answer:

See below

Step-by-step explanation:

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

Recall

$\dfrac{d}{dx}\tan x=\sec^2$

Using the chain rule

$\dfrac{dy}{dx}= \dfrac{dy}{du} \dfrac{du}{dx}$

such that $u = \tan x$

we can get a general formulation for

$y = \tan^n x$

Considering the power rule

$\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}$

we have

$\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x$

therefore,

$\dfrac{d}{dx}\tan^3 x=3\tan^2x \sec^2x$

Now, once

$\sec^2 x - 1= \tan^2x$

we have

$3\tan^2x \sec^2x = 3(\sec^2 x - 1) \sec^2x = 3\sec^4x-3\sec^2x$

Hence, we showed

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

================================================

For the integration,

$\int \sec^4 x\, dx$

considering the previous part, we will use the identity

$\boxed{\sec^2 x - 1= \tan^2x}$

thus

$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$

and

$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx$

Considering $u = \tan x$

and then $du=\sec^2x\ dx$

we have

$\int u^2 \, du = \dfrac{u^3}{3}+C$

Therefore,

$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$

$\boxed{\int \sec^4 x\, dx = \dfrac{\tan^3 x}{3}+\tan x + C }$

6 0

3 years ago

Please help me solve this

Rashid [163]

Answer:

??????????????

Step-by-step explanation:

,nooooooo

8 0

3 years ago