(8-3)²-(5-2)²
Follow the PEMDAS
P-Parenthesis
E-Exponents
M/D-Multiplication/Division
A/S-Addition/Subtraction
clear the numerals in the parenthesis:
(5)²-(3)²
Now do the exponents:
=25-9
=16
Its just: (5)²-(3)²
9514 1404 393
Answer:
x = 7
Step-by-step explanation:
You solve a linear equation by putting the variable on one side of the equal sign and a constant on the other side. Here, variables and constants are on both sides of the equal sign, so you need to separate them.
The basic idea is that you add the opposite of any term you don't want. Whenever you perform any operation (like "add"), <em>you must do it to both sides of the equation</em>.
We observe that x-terms have coefficients of 10 and 9. We choose to add the opposite of 9x to both sides:
10 -9x -5 = 9x -9x +2
x -5 = 2 . . . . simplify
Now, we still have -5 on the left, where we don't want it. So, we add its opposite (+5) to both sides:
x -5 +5 = 2 +5
x = 7 . . . . simplify
The solution is x = 7.
_____
<em>Additional comment</em>
If we were to end up with an x-coefficient other than 1, we would divide both sides of the equation by that coefficient. This will leave the x-term with a coefficient of 1.
Answer
Step-by-step explanation:
The area would be length times width. x(4x+8) =96
4x^2+8x = 96
4x^2+8x-96=0. We need to factor.
It would be factored to (4x-16)(x+6) = 0
x would be -6 (x+6=0)
or x would be 4 (4x-16=0)
Why would the width be negative?!? So the answer is 4
Answer:
y=5x/2 + 9
Step-by-step explanation:
Check the above picture for workings
Answer:
a) 0.82
b) 0.18
Step-by-step explanation:
We are given that
P(F)=0.69
P(R)=0.42
P(F and R)=0.29.
a)
P(course has a final exam or a research paper)=P(F or R)=?
P(F or R)=P(F)+P(R)- P(F and R)
P(F or R)=0.69+0.42-0.29
P(F or R)=1.11-0.29
P(F or R)=0.82.
Thus, the the probability that a course has a final exam or a research paper is 0.82.
b)
P( NEITHER of two requirements)=P(F' and R')=?
According to De Morgan's law
P(A' and B')=[P(A or B)]'
P(A' and B')=1-P(A or B)
P(A' and B')=1-0.82
P(A' and B')=0.18
Thus, the probability that a course has NEITHER of these two requirements is 0.18.
