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3 years ago

(-5,-3) (5,-4) slope =

Mathematics

1 answer:

bija089 [108]3 years ago

7 0

Answer:

Your slope would be -1/10

Step-by-step explanation:

To find the slope from two points, you would use the formula y2-y1/x2-x1

In this case, the equation would look like -4-(-3)/5-(-5)

-4 - (-3) = -1

5 - (-5) = 10

-1/10 = -1/10

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3 years ago

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3 years ago

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7. The bottom of a circular swimming pool

mote1985 [20]

<h2>Hello!</h2>

The answer is:

1017.88 square feet are blue.

To calculate how many square feet are blue, we need to calculate the area of the bottom of the circular swimming pool.

We need to remember that we can calculate the area of any circle using the following formula:

$Area_{circle}=\pi radius^{2}$

Now, we know that the diameter of the bottom is equal to 36 feet, so, we can calculate the radius of the bottom using the following formula:

$diameter=2radius\\\\radius=\frac{diameter}{2}$

So,

$radius=\frac{36feet}{2}=18feet$

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$Area_{bottom}=\pi radius^{2}=\pi 18ft^{2}=1017.88ft^{2}$

Hence, we have that 1017.88 square feet are blue.

Have a nice day!

4 0

3 years ago

Kain graphs the hyperbola (y+2)^2/64 − (x+5)^2/36 = 1 . How does he proceed? Drag a value, phrase, equation, or coordinates in t

cupoosta [38]

Answer:

1st box: (-5,-2)

2nd box: up and down

3rd box: 8

4th box: +-4/3

5th box: y+2 = +-4/3 (x+ 5)

Step-by-step explanation:

Just took the k12 unit test, hope this helps!

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3 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 $9\text{ln}|x|+2\sqrt{x}+x+C$ .

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3 years ago