C = pi * d
a = pi * (d/2)^2
I would first add 5 and 2 together, then multiply that by 3. Finally you would subtract the 1. After solving for what is inside the parentheses you would multiply it by 2 for your final answer.
2•(3(5+2)-1)
2•(3(7)-1)
2•(21-1)
2•(20)
40
Given,
The sum of three integers is 92
So,
Let,
The first integer be "x"
The second integer be "y"
The third integer be "z"
Now,
According to the question,
y = 3x ..............equation (1)
z = 2x - 10 .............. equation (2)
x + y + z = 92 ..............equation (3)
Now,
Substituting the value of "y" and "z" from equation (1) and (2), we get,
x + (3x) + (2x - 10) = 92
x + 3x + 2x - 10 = 92
6x - 10 = 92
6x = 92 + 10
x = 102 / 6
x = 17
Now,
substituting the value of "x" in equation (1)
y = 3 (17)
y = 51
Now,
Substituting the value of "x" in equation (2),
z = 2 (17 ) - 10
z = 34 - 10
z = 24
So, the numbers are 17, 51 and 24
NEVER HATE MATH!!!
