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2 years ago

Solve each system by substitution. 5) 5x + 3y =-23 y = 4

Mathematics

1 answer:

docker41 [41]2 years ago

3 0

Answer:

(- 7, 4 )

Step-by-step explanation:

Given the 2 equations

5x + 3y = - 23 → (1)

y = 4 → (2)

Substitute y = 4 directly into (1_

5x + 3(4) = - 23, that is

5x + 12 = - 23 ( subtract 12 from both sides )

5x = - 35 ( divide both sides by 5 )

x = - 7

Solution is (- 7, 4 )

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Help me please . I am stuck

Radda [10]

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Step-by-step explanation:

I don;t know how I can use the binomial squares pattern on Brainly, but I can definitely guide you through the problem.

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3 years ago

Find the length of the hypotenuse of a right triangle if a = 3 and b = 4.

alukav5142 [94]

The answer is the hypotenuse has a length of 5.

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2 years ago

Tabitha bought an antique model car priced at $72. She also had to pay 5% sales tax. What was the total amount she paid?

Pani-rosa [81]

Answer:

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Step-by-step explanation:

Multiply the model car's price by 105% (or 1.05) to find the total price.

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5 0

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

1 , 3 , 6 , 10 , 15

Margarita [4]

As we can see that common difference is increasing to one more digit after every term

so it means

a21 = 231

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2 years ago

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