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2 years ago

Factor completely.

Mathematics

2 answers:

Tomtit [17]2 years ago

5 0

Answer:

Step-by-step explanation:

y² - 12y + 32

consider the product of factors of the constant term (+ 32) which sum to give the coefficient of the y- term (- 12)

the factors are - 4 and - 8 , since

- 4 × - 8 = + 32 and - 4 - 8 = - 12 , then

y² - 12y + 32 = (y - 4)(y - 8)

inna [77]2 years ago

5 0

Answer:

B. (y-4)(y-8)

Step-by-step explanation:

Given equation: y² - 12y + 32. Essentially, we are going to find two numbers that multiply to 32 and add up to -12.

1) Pull out the factors of 32 and find the pair that multiply to 32 and add up -12.

They are: -8 and -4.

2) Rewrite the b of the original quadratic equation as a sum, then factorise by grouping.

y² - 8y - 4y +32

y(y - 8) - 4(y - 8)

(y - 4)(y -8)

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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:

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

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

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