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AleksAgata [21]
3 years ago
15

Write the equation of the line that passes through the given point and has the given slope:

Mathematics
2 answers:
My name is Ann [436]3 years ago
8 0

Answer:

y=-x-9

Step-by-step explanation:

y-y1=m(x-x1)

y-0=-1(x-(-9))

y=-1(x+9)

y=-x-9

bekas [8.4K]3 years ago
3 0

Answer: y + x + 9 = 0

Step-by-step explanation:

equation of a line with the following values is

y - y1

-------   = -1

x - x1

enter the values of the coordinate x1 and y1 into the formula above

y - 0

-------    = -1

x - (-9)

y - 0

-------    = -1

x + 9

now cross multiply to have it in a linear form

y - 0 = -1( x + 9 )

y - 0 = -x - 9

y + x = -9

y + x + 9 =0

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Two angles of a triangle are the same size. The third angle is 3 times as large
Gnoma [55]
Use x to represent the first two angles. If the third angle is 3 times as large as the first, it can be represented as 3x. So, your angles are x, x, and 3x. These need to all add to 180 degrees, since it’s a triangle.

x + x + 3x = 180 —> Simplify the left side.
5x = 180 —> Divide by 5.
x = 36

So the first two angles are 36 degrees, and the third angle is 3 times that, which is 108 degrees.
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3 years ago
Find the area of a circle with a diameter of 8cm.
Elza [17]

Answer:

50.24

Step-by-step explanation:

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4) Part A draw half of a trapezoid
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Whats the value of x????
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Calculus Problem
Roman55 [17]

The two parabolas intersect for

8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2

and so the base of each solid is the set

B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, |x^2-(8-x^2)| = 2|x^2-4|. But since -2 ≤ x ≤ 2, this reduces to 2(x^2-4).

a. Square cross sections will contribute a volume of

\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x

where ∆x is the thickness of the section. Then the volume would be

\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x

We end up with the same integral as before except for the leading constant:

\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx

Using the result of part (a), the volume is

\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x

and using the result of part (a) again, the volume is

\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}

7 0
2 years ago
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