14 x 2000
Basic multiplication fact:
14 x 2=28
Total number of zeros=3
So, 14 x 2000=28,000
F(x) + k - Moves the graph k units up.
k f(x) stretches the graph parallel to y-axis by a facor k
f (kx) stretches the graph by a factor 1/k parallel to x-axis
f(x + k) moves the graph 3 units to the left.
For k negative the first one moves it k units down
for second transform negative does same transfoormation but also reflects the graph in the x axis
For the third transform negative k :- same as above but also reflects in y axis
4th transform - negative k moves graph k units to the right
It’s 192% because you just need to remove all the points :) hope i helped
6m Divided by 2.7 Seconds
6/2.7=<span>2.22222222222
2.22222222 Meters per second is end velocity
the change in velocity is Start velocity (0m/s) - end velocity (2.222222222m/s)
So the Change in velocity is 2.222222222m/s
To find acceleration it is Change in velocity (2.2222222m/s)/Time (2.7m/s)
2.222222222222/2.7=</span><span>0.82304526749
acceleration=</span><span>0.82304526749 Meters per second Squared
This also would make sense as we already know acceleration is less on the moon.
Answer=</span>0.82304526749m/s ²<span> </span>
Answer:
Therefore, the volume of the cone is V=4π.
Step-by-step explanation:
From task we have a circular cone with radius 2 m and height 3 m. We use the disk method to find the volume of this cone.
We have the formula:
We know that r=2 and h=3, and we get:
![V=\int_0^3\pi \cdot \left(\frac{2}{3}x\right)^2\, dx\\\\V=\int_0^3 \pi \frac{4}{9}x^2\, dx\\\\V= \frac{4\pi}{9} \int_0^3 x^2\, dx\\\\V= \frac{4\pi}{9} \left[\frac{x^3}{3}\right]_0^3\, dx\\\\V= \frac{4\pi}{9}\cdot 9\\\\V=4\pi](https://tex.z-dn.net/?f=V%3D%5Cint_0%5E3%5Cpi%20%5Ccdot%20%5Cleft%28%5Cfrac%7B2%7D%7B3%7Dx%5Cright%29%5E2%5C%2C%20dx%5C%5C%5C%5CV%3D%5Cint_0%5E3%20%5Cpi%20%5Cfrac%7B4%7D%7B9%7Dx%5E2%5C%2C%20dx%5C%5C%5C%5CV%3D%20%5Cfrac%7B4%5Cpi%7D%7B9%7D%20%5Cint_0%5E3%20x%5E2%5C%2C%20dx%5C%5C%5C%5CV%3D%20%5Cfrac%7B4%5Cpi%7D%7B9%7D%20%5Cleft%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%5Cright%5D_0%5E3%5C%2C%20dx%5C%5C%5C%5CV%3D%20%5Cfrac%7B4%5Cpi%7D%7B9%7D%5Ccdot%209%5C%5C%5C%5CV%3D4%5Cpi)
Therefore, the volume of the cone is V=4π.
