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3 years ago

Rewrite the equation by completing the square X^2+6x+9=0 (X+?)^2=?

Mathematics

1 answer:

Cerrena [4.2K]3 years ago

5 0

Answer:

x^2 + 6x +9 = 0

(x+3)^2 = 0

(x+3)(x+3)= 0

Step-by-step explanation:

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3 years ago

Who can solve this <br>2xy if x=5 and y=-3

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Answer:

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Step-by-step explanation:

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

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For the function f(x)=5x^2-7x, what is f(-3)?

Ainat [17]

Answer:

Step-by-step explanation:

f(x)=5x^2-7x

substitute x to 3

f(3)=5(3)^2-7(3)

f(3)=5(9)-21

f(3)=45-21

f(3)=24

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3 years ago