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madreJ [45]
3 years ago
5

What is the slope of the line that passes through the points (7, -2 ) and ( -1, -5)

Mathematics
1 answer:
VLD [36.1K]3 years ago
4 0

A

to calculate the slope m use the gradient formula

m = ( y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (7, - 2 ) and (x₂, y₂ ) = (- 1, - 5 )

m = \frac{-5+2}{-1-7} = \frac{-3}{-8} = \frac{3}{8} → A


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y= 3x+ 1

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Express 82 kilometers per hour in miles per hour
mezya [45]

Step-by-step explanation:

a)82km into miles

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6 0
3 years ago
PLEASE HELP I CANT UNDERSTAND I NEED HELP
Vika [28.1K]

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Count houw many boxes the linegoes over

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
ПОЖАЛУЙСТА ПОМОГИТЕ РЕШИТЬ ДАЮ 23 ОЧКА!!!
posledela

(Простите, пожалуйста, мой английский. Русский не мой родной язык. Надеюсь, у вас есть способ перевести это решение. Если нет, возможно, прилагаемое изображение объяснит достаточно.)

Use the shell method. Each shell has a height of 3 - 3/4 <em>y</em> ², radius <em>y</em>, and thickness ∆<em>y</em>, thus contributing an area of 2<em>π</em> <em>y</em> (3 - 3/4 <em>y</em> ²). The total volume of the solid is going to be the sum of infinitely many such shells with 0 ≤ <em>y</em> ≤ 2, thus given by the integral

\displaystyle 2\pi \int_0^2 y \left(3-\frac34 y^2\right) \,\mathrm dy = 2\pi \left(\frac32 y^2 - \frac3{16} y^4\right)\bigg|_0^2 = 6\pi

Or use the disk method. (In the attachment, assume the height is very small.) Each disk has a radius of √(4/3 <em>x</em>), thus contributing an area of <em>π</em> (√(4/3 <em>x</em>))² = 4<em>π</em>/3 <em>x</em>. The total volume of the solid is the sum of infinitely many such disks with 0 ≤ <em>x</em> ≤ 3, or by the integral

\displaystyle \pi \int_0^3 \left(\sqrt{\frac43x}\right)^2 \,\mathrm dx = \frac{2\pi}3 x^2\bigg|_0^3 = 6\pi

Using either method, the volume is 6<em>π</em> ≈ 18,85. I do not know why your textbook gives a solution of 90,43. Perhaps I've misunderstood what it is you're supposed to calculate? On the other hand, textbooks are known to have typographical errors from time to time...

8 0
3 years ago
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