We Can Let "A Number" Be M.
M Divided By 3 Less Than Itself Gives Quotient Of 8/5.
M/(M-3) = 8/5. Lets Solve.
So, What Number Can Have 3 Subtracted From It To Get Five, But Also Be Equal To Eight?
Eight Of Course!!!
So, We Know That The Number Equals 8.
1) given x^4 + 95x^2 - 500
2) split in two factors with common factor term x^2: (x^2 + )(x^2 - )
3) find two numbers that add up 95 and their product is - 500:
=> 100*(-5) = - 500 and 100 - 5 = 95
=> (x^2 + 100)(x^2 - 5)
4) factor x^2 - 5 = (x + √5) (x - √5)
5) write the prime factors: (x^2 + 100) (x + √5) (x -√5)
6) find the solutions:
x^2 + 100 = 0 => not possible
x + √5 = 0 => x = - √5
x - √5 = 0 => x = √5
Answer: x = √5 and x = - √5
4n+7−(7n−8)
=4n+7+−1(7n−8)
=4n+7+−1(7n)+(−1)(−8)
=4n+7+−7n+8
Combine Like Terms
=4n+7+−7n+8
=(4n+−7n)+(7+8)
=−3n+15
Acc. to midpoint theorem,
15={(2x+9) + (4x-15)}/2
⇒ 30=2x+9+4x-15
⇒ 30-9+15=6x
⇒ 36/6=x
⇒6=x
WZ= 4x-15
=4*6-15
=24-15=9