Answer:
the volume of the sphere will be 4 times the volume of the cone;
Step-by-step explanation:
<u>The question is on volume comparison</u>
Volume of a sphere=4/3
×r³
Volume of a cone=
/3×r²h
where r is the radius and h is the height
<u>Apply the condition</u>
If r=h=1 unit
Then volume of sphere will be= 4/3 ×
×1³ = 4/3![\pi](https://tex.z-dn.net/?f=%5Cpi)
And volume of the cone will be=
/3 ×1²×1 =
/3
We can see the volume of the sphere will be 4 times the volume of the cone;
4/3
= 4×
/3
F(2) means when x=2, what is y
assuming each grid is 1 unit
when x=2, y=-1
f(2)=-1
f(1/2) is where y is when x=1/2
that is at y=1
f(1/2)=1
f(-2) is where y is when x=-2
that is at y=2
f(-2)=2
No.
4p - 5 = 16
Isolate the variable and the constant.
4p = 21
p = 21/4
Quotient means divide.....increased means add
n/4 + 5 <===
Answer:
![\frac{x^3-a^3}{0}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E3-a%5E3%7D%7B0%7D)
Step-by-step explanation:
![\frac{x^4-a^4}{x-a}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E4-a%5E4%7D%7Bx-a%7D)
divide by x-a
![\frac{x^3-a^3}{0}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E3-a%5E3%7D%7B0%7D)