Answer:
6^-2
6^3/6^5
6^-9 *6^ 7
Step-by-step explanation:
We want the fractions equal to 1/36
We know that a^-b = 1/a^b
We also know that a^b / a^c = a^(b-c)
We also know that a^b * a^c = a^(b+c)
3^ -6 = 1/3^6 = 1/729
6^-2 = 1/6^2 = 1/36
6^3/6^5 = 6^(3-5) = 6^-2 = 1/6^2 = 1/36
6^2 / 6^-1 = 6^(2- -1) = 6^3 = 216
6*6^-2 = 6^(1-2) = 6^-1 = 1/6
6^-9 *6^ 7 = 6^(-9+7) = 6^ -2 = 1/6^2 = 1/36
Answer:
<em>96π units²</em>
Step-by-step explanation:
Find the diagram attached
Area of a sector is expressed as;
Area of a sector = θ/2π * πr²
Given
θ = 3π/4
r = 16
Substitute into the formula
area of the sector = (3π/4)/2π * π(16)²
area of the sector = 3π/8π * 256π
area of the sector = 3/8 * 256π
area of the sector = 3 * 32π
<em>area of the sector =96π units²</em>
Answer:
A and D
Step-by-step explanation:
Here, we shall be evaluating the validity of the statements;
A. Yes, A is true
There are four even numbers 2,4,6 and 8 and 4 odd number 1,3,5,7; The landing should be equal at 125 each
B. This is wrong
It is supposed to land half of the number of time s which is half of 250 and that is 125
C.This is wrong
The numbers greater than 4 are 5,6,7,8
Now, the probability should be 4/8 = 1/2 and that is 50%
D. This is correct
Number of times we have a landing on odd numbers is 250-135 = 115
The experimental probability of landing on an odd number is thus 115/250 = 0.46 which is 46%
Answer:
Step-by-step explanation:
1 ) Given that
For a non homogeneous part 
, we assume the particular solution is
2 ) Given that
For a non homogeneous part 
, we assume the particular solution is
3 ) Given that
y′′ + 4y′ + 20y = −3sin(2x)
For a non homogeneous part −3sin(2x) , we assume the particular solution is
4 ) Given that
y′′ − 2y′ − 15y = 3xcos(2x)
For a non homogeneous part 3xcos(2x) , we assume the particular solution is
