Um, we see that the slice created the triangle, and the formula to find the area of the triangle:
1/2×B×h where B represent Base and h represent height.
In this case, Base=5 cm
Height=11 cm
A=1/2×11×5
A=5.5×5
A=27.5 square cm. As a result, the area of the resulting two-dimensional cross section is 27.5 square cm. Hope it help!
Answer:
Step-by-step explanation:
Given
Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height
Suppose they have same mass and radius
time Period is given by
,where h=height of release
a=acceleration

Where I=moment of inertia
a for hoop


a for Uniform solid cylinder


a for spherical shell


a for Uniform Solid


time taken will be inversely proportional to the square root of acceleration




thus first one to reach is Solid Sphere
second is Uniform solid cylinder
third is Spherical Shell
Fourth is hoop
The answer is definitely 30 :) thanks for the points!
So first you plug 6 into the equation which is 4(6)-10.
Using PEMDAS, you solve what’s in the parenthesis first and you get 24-10.
24-10=14
f(6)=14
Your fully factored expression should be 2*(8a+5).