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Hitman42 [59]
2 years ago
10

Graph a line with a slope of -3,that contains the point (4, -2)

Mathematics
1 answer:
Phantasy [73]2 years ago
6 0

Answer:

Hi! The equation used to graph a line with a slope of -3 that contains the point (4,-2) is (-3x+10).

The graph would look like this (if it doesn't show up please let me know)

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kipiarov [429]

we know that

The unit rate is $0.65 per foot

so

To find out the cost, multiply the unit rate by the length of the rope

so

0.65*(10.8)=$7.02

therefore

<h2>The cost is $7.02</h2>
3 0
1 year ago
Write the equation of the line passing through points (2,-2) and (1,0) in slope intercept form.
katrin2010 [14]

Answer:

y=-2x+2

Step-by-step explanation:

hope it helps!

8 0
3 years ago
Read 2 more answers
The circumference of a circle is 75.36 cm. What is the area of the circle?
Anuta_ua [19.1K]
The area of a circle, given the circumference, is found using this formula:

A = C²/4π

Therefore,
A ≈ 451.93 cm²
7 0
3 years ago
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Given f(x)=x-10. what is f(-7)
Vlad [161]
f( - 7) = ( - 7) - 10 \\ = -1 7
answer is -17
5 0
3 years ago
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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