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2 years ago

Solve the system by the elimination method.

Mathematics

1 answer:

REY [17]2 years ago

8 0

Answer: C is correct, 4x=8

Step-by-step explanation:

You will add the two equations...

4x-8=0

Now you need to move the -8 over, which reverses its sign to positive.

4x=8

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this week Rajiv's puppy weighs 8 kilograms. Last week the puppy weighed 6.5 kilograms how much mass in grams did the puppy gains

Olegator [25]

Answer:

1,500 G

Step-by-step explanation:

1 Kilogram (Kg) = 1,000 Grams (G)

8-6.5= 1.5

1.5x1,000=1,500

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2 years ago

What is the answer to -26-(-49)?

Radda [10]

Answer:

Step-by-step explanation:

-26-(-49)=-26+49=23

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2 years ago

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How to solve 2-7d=15

melomori [17]

Answer:-7d=13

7d=-13

d=-13/7

Step-by-step explanation:

Hope this helps!

Make sure to apply same operations to both sides.

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2 years ago

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PLS help timed! which graph represents the solution of y<x2-1 and x>y2-3?

miv72 [106K]

Answer:

y minus 3 = 2x plus 1

Step-by-step explanation:

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2 years ago

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Find the general indefinite integral. (Use C for the constant of integration. Remember to use absolute values where appropriate.

Maru [420]

Answer:

$9\text{ln}|x|+2\sqrt{x}+x+C$

Step-by-step explanation:

We have been an integral $\int \frac{9+\sqrt{x}+x}{x}dx$ . We are asked to find the general solution for the given indefinite integral.

We can rewrite our given integral as:

$\int \frac{9}{x}+\frac{\sqrt{x}}{x}+\frac{x}{x}dx$

$\int \frac{9}{x}+\frac{1}{\sqrt{x}}+1dx$

Now, we will apply the sum rule of integrals as:

$\int \frac{9}{x}dx+\int \frac{1}{\sqrt{x}}dx+\int 1dx$

$9\int \frac{1}{x}dx+\int x^{-\frac{1}{2}}dx+\int 1dx$

Using common integral $\int \frac{1}{x}dx=\text{ln}|x|$ , we will get:

$9\text{ln}|x|+\int x^{-\frac{1}{2}}dx+\int 1dx$

Now, we will use power rule of integrals as:

$9\text{ln}|x|+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\int 1dx$

$9\text{ln}|x|+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+\int 1dx$

$9\text{ln}|x|+2x^{\frac{1}{2}}+\int 1dx$

$9\text{ln}|x|+2\sqrt{x}+\int 1dx$

We know that integral of a constant is equal to constant times x, so integral of 1 would be x.

$9\text{ln}|x|+2\sqrt{x}+x+C$

Therefore, our required integral would be $9\text{ln}|x|+2\sqrt{x}+x+C$ .

4 0

2 years ago