SOLUTION;
![\sqrt[]{-36}\text{ = }\sqrt[]{(36)(-1)}\text{ = }\sqrt[]{36}\text{ x }\sqrt[]{-1\text{ }}\text{ = 6i}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B-36%7D%5Ctext%7B%20%3D%20%7D%5Csqrt%5B%5D%7B%2836%29%28-1%29%7D%5Ctext%7B%20%3D%20%7D%5Csqrt%5B%5D%7B36%7D%5Ctext%7B%20x%20%7D%5Csqrt%5B%5D%7B-1%5Ctext%7B%20%7D%7D%5Ctext%7B%20%3D%206i%7D)
Recall that the square root of the negative one is "i" meaning that it is a complex number and not a real number.
A
evaluate f(5) and f(2)
f(5) = 5m + b and f(2) = 2m + b, hence
f(5) - f(2) = 5m + b - 2m - b = 3m
the expression simplifies to
= 2 ( cross- multiply )
3m = 6 ( divide both sides by 3 )
m = 2 → A
The equation is 
.
1/2 is equivalent to 4/8, and 2 1/8 is equal to 17/8. (two full sets of 8 + 1 = 8 + 8 + 1, or 17)
The new equation is 
. Subtract the numerators.
17 - 4 = 13 or 13/8
The improper fraction 
simplifies to 
.
<h2>Answer:</h2>
Hope this helps :)
Answer:
i need more info
Step-by-step explanation:
Answer:
The value of A is 5
Step-by-step explanation:
- The number is divisible by 3 if the sum of its digits is a number
divisible by 3
- Ex: 126 is divisible by 3 because the sum of its digits = 1 + 2 + 3 = 6
and 6 is divisible by 3
- The number is divisible by 5 if its ones digit is zero or 5
- Ex: 675 is divisible by 5 because its ones digit is 5
890 is divisible by 5 because its ones digit is 0
- We are looking for the value of A in the 4-digit number 3A5A which
makes the number divisible by both 3 and 5
∵ A is in the ones position
∴ A must be zero or 5
- Let us try A = 0
∵ A = 0
∴ The number is 3050
∵ The sum of the digits of the number = 3 + 0 + 5 + 0 = 8
∵ 8 is not divisible by 3
∴ 3050 is not divisible by both 3 and 5
∴ A can not be zero
- Let us try A = 5
∵ A = 5
∴ The number is 3555
∵ The sum of the digits of the number = 3 + 5 + 5 + 5 = 18
∵ 18 is divisible by 3
∴ 3555 is divisible by both 3 and 5
∴ A must be equal 5
* <em>The value of A is 5</em>
