Answer:
Step-by-step Explanation:
==>Given:
Dimensions of a rectangular prism are expressed as follow:
Volume (V) = 15x² + x + 2
Height (h) = x²
==>Required:
Expression of the Base area (B)
==>Solution:
Volume (V) = Base (B) × Height (h)
15x² + x + 2 = B × x²
Divide both sides by x²
![\frac{15x² + x + 2}{x²} = B[tex]Base (B) = /frac{15x² + 1 + 2}{x}](https://tex.z-dn.net/?f=%5Cfrac%7B15x%C2%B2%20%2B%20x%20%2B%202%7D%7Bx%C2%B2%7D%20%3D%20B%3C%2Fp%3E%3Cp%3E%5Btex%5DBase%20%28B%29%20%3D%20%2Ffrac%7B15x%C2%B2%20%2B%201%20%2B%202%7D%7Bx%7D)
8.9 because I used a calculator
Answer:
5x + 3
Step-by-step explanation:
Since we can see DC is the median of Triangle ABC
So it breaks the into two equal parts
Given
AD = 2x + 5
BD = 3x - 2
Now
AB = AB + BD
= 2x + 5 + 3x - 2
= 5x + 3
Hope it will help :)❤
Step-by-step explanation:
Let the height above which the ball is released be H
This problem can be tackled using geometric progression.
The nth term of a Geometric progression is given by the above, where n is the term index, a is the first term and the sum for such a progression up to the Nth term is
To find the total distance travel one has to sum over up to n=3. But there is little subtle point here. For the first bounce ( n=1 ), the ball has only travel H and not 2H. For subsequent bounces ( n=2,3,4,5...... ), the distance travel is 2×(3/4)n×H
a=2H..........r=3/4
However we have to subtract H because up to the first bounce, the ball only travel H instead of 2H
Therefore the total distance travel up to the Nth bounce is
For N=3 one obtains
D=3.625H
Answer:
The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. ... No; The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. ANSWER: No; 9.
