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2 years ago

Find the slope between (0, 0) and (3, 4)

Mathematics

1 answer:

morpeh [17]2 years ago

5 0

Answer:

Slope = 4/3

Step-by-step explanation:

Let's find the slope between your two points.

(0, 0) ; (3, 4)

(x1, y1) = (0, 0)

(x2, y2) = (3, 4)

Use the slope formula:

m = y2 − y1 / x2 − x1

= 4 − 0 / 3 − 0

= 4 / 3

Answer:

m = 4 / 3

hope this helps!! p.s. i really need brainliest :)

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9 of the students in Sarah's grade like to watch football and 12 like to watch figure skating. 7 students like to watch both foo

madreJ [45]

Answer:

Step-by-step explanation:

Total for football: 9 +7

Total for figure skating: 12 +7

NEW Total for football: 16

NEW Total for figure skating: 19

The question says that 9 people like football

The question says that 12 people like to watch figure skating

The question also says that 7 people like to watch both, and because of that, you should add 7 to both 9 and 12, but keep the new numbers separated from each other. So once we know the sum of 9 +7 is 16, and 12 + 7 is 19, we can subtract the two numbers, and get 3 as the answer. Because you are counting the number of times that you have to travel up or down to the other number, and so if you are on a set of stairs that go up to the 16 level and you are on the 14th level, you need to travel up 2 sets of stairs to get to level 16.

Hope that example helps and can help in the future with other problems!

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Step-by-step explanation:

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3 years ago

The product of 718 and 82

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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

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

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2 years ago