<em>3</em><em>9</em><em>></em><em>3</em><em>8</em>
<em>1</em><em>6</em><em>=</em><em>1</em><em>6</em>
<em>1</em><em>1</em><em><</em><em>3</em><em>1</em><em>5</em>
<em>5</em><em>></em><em>4</em>
<em>Hope </em><em>this </em><em>helps </em><em>you </em><em>mate </em>
<em>~♥~</em><em>♪☆\(^0^\) ♪(/^-^)/☆</em><em>♪☆\(^0^\) ♪(/^-^)/☆</em><em>♪\(*^▽^*)/\(*^▽^*)/</em><em>♪☆\(^0^\) ♪(/^-^)/☆</em><em>♪ ♬ ヾ(´︶`♡)ノ ♬ ♪</em>
Y=1 1/3x + 6
Slope = y2-y1/x2-x1
2- -2 = 4
-3 - -6 = 3
4/3 = 1 1/3
<u>Answer:</u>
5 miles
<u>Step-by-step explanation:</u>
15 × ? = 25
3 × ? = ?
15 × 1 and 2/3 = 25
3 × 1 and 2/3 = ?
3 × 1 and 2/3 = 5
Your welcome and Rate as Brainliest
Answer:
The radius is increasing at a rate of 62832 cubic millimeters per second when the diameter is of 100 mm.
Step-by-step explanation:
Volume of a sphere:
The volume of a sphere of radius r is given by:
![V = \frac{4\pi r^3}{3}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B4%5Cpi%20r%5E3%7D%7B3%7D)
How fast is the volume increasing:
To find this, we have to differentiate the variables of the problem, which are V and r, implicitly in function of time. So
![\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%20%3D%204%5Cpi%20r%5E2%5Cfrac%7Bdr%7D%7Bdt%7D)
The radius of a sphere is increasing at a rate of 2 mm/s.
This means that ![\frac{dr}{dt} = 2](https://tex.z-dn.net/?f=%5Cfrac%7Bdr%7D%7Bdt%7D%20%3D%202)
How fast is the volume increasing (in mm3/s) when the diameter is 100 mm?
Radius is half the diameter, so ![r = \frac{100}{2} = 50](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B100%7D%7B2%7D%20%3D%2050)
Then
![\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt} = 4\pi (50)^2(2) = 62832](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%20%3D%204%5Cpi%20r%5E2%5Cfrac%7Bdr%7D%7Bdt%7D%20%3D%204%5Cpi%20%2850%29%5E2%282%29%20%3D%2062832)
The radius is increasing at a rate of 62832 cubic millimeters per second when the diameter is of 100 mm.
Answer:
Step-by-step explanation: