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3 years ago

You are an administrative assistant. Your boss has

1 answer:

7 0

Option A: 156 + 179 + 25 + 25 = $385
Option B: 334 + 15 + 15 = $364

Option B is cheaper. 10% of $364 is $36.40. That is what you would save with the third option.

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Which is the correct input-output table for the function g(x)=-1/2(4x+6)

Sonbull [250]

Answer: The third choice.

Step-by-step explanation: To find the first output, start with your first input. In this case, it's -2, so your equation would be g(-2)= -1/2(4(-2)+6). Start by solving inside the parenthesis. 4x-2=-8 and -8+6=-2. -1/2x-2= 1, so your first output should be 1. The only choice that has this is the third one, so that is your answer. Hope I could help :)

4 0

3 years ago

Simplify (3x-5x)+(5x+1)

Lemur [1.5K]

First do
3x + -5x = -2x
Then do
-2x + 5x = 3x
Lastly do
3x+1

6 0

3 years ago

Help me complete it fast pls

andrew11 [14]

Answer:

AC < AB

Step-by-step explanation:

We can see just by looking at it, they are not the same length.

8 0

2 years ago

Solve this system of equations using substitution, show your work (ANSWER QUICK PLEASE)

melisa1 [442]

Answer:

$\boxed{\sf{y=-50 \quad x=3}}$

Step-by-step explanation:

Isolate the term of x and y from one side of the equation.

First, you have to substitute.

$\sf{y=-15x-5=11x-15x-5=-17}$

Then, you solve.

$\sf{-4x-5=-17}$

Add by 5 from both sides.

-4x-5+5=-17+5

Solve.

-17+5=-12

-4x=-12

Divide by -4 from both sides.

-4x/-4=-12/-4

Solve.

-12/-4=3

y=-15*3-5

Solve.

PEMDAS stands for:

Parenthesis
Exponents
Multiply
Divide
Add
Subtract

15*3=45

Rewrite the problem down.

y=-45-5

Solve.

<u>Therefore, the correct answer is y=-50 and x=3.</u>

I hope this helps you! Let me know if my answer is wrong or not.

3 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Crm%5Cfrac%7Bd%7D%7Bdx%7D%20%20%20%5Cleft%20%28%20%5Cbigg%28%20%5Cint_%7B1%7D%5E%7B%20%7B

Margaret [11]

Applying the product rule gives

$\displaystyle \frac{d}{dx}\int_1^{x^2}\frac{2t}{1+t^2}\,dt \times \int_1^{\ln(x)}\frac{dt}{(1+t)^2} + \int_1^{x^2}\frac{2t}{1+t^2}\,dt \times \frac{d}{dx}\int_1^{\ln(x)}\frac{dt}{(1+t)^2}$

Use the fundamental theorem of calculus to compute the remaining derivatives.

$\displaystyle \frac{4x^3}{1+x^4} \int_1^{\ln(x)}\frac{dt}{(1+t)^2} + \frac{1}{x(1+\ln(x))^2}\int_1^{x^2}\frac{2t}{1+t^2}\,dt$

The remaining integrals are

$\displaystyle \int_1^{\ln(x)}\frac{dt}{(1+t)^2} = -\frac1{1+t}\bigg|_1^{\ln(x)} = \frac12-\frac1{1+\ln(x)}$

$\displaystyle \int_1^{x^2}\frac{2t}{1+t^2}\,dt=\int_1^{x^2}\frac{d(1+t^2)}{1+t^2}=\ln|1+t^2|\bigg|_1^{x^2}=\ln(1+x^4)-\ln(2) = \ln\left(\frac{1+x^4}2\right)$

and so the overall derivative is

$\displaystyle \frac{4x^3}{1+x^4} \left(\frac12-\frac1{1+\ln(x)}\right) + \frac{1}{x(1+\ln(x))^2} \ln\left(\frac{1+x^4}2\right)$

which could be simplified further.

5 0

2 years ago