Answer:
x=4
Step-by-step explanation:
all you need to do is the 3rd root of 64
which is 4.
Answer:
5x^2+22x-12 x cannot be -5, -4, -2
(x+5)(x+4)(x+2)
Step-by-step explanation:
In order to solve this, your denominator must be the same. Let's start by writing out the two different quadratic formulas:
x^2 + 6x + 8 <-- This should factor out to (x+4)(x+2)
x^2 + 7x + 10 <-- This should factor out to (x+5)(x+2)
Now that you have factored out the two quadratics, plug them into the equation.
5x - 3
(x+4)(x+2) (x+5)(x+2)
Now as we know, -2 cannot be x because it will turn the entire equation undefined. Multiple top and bottom with (x+5) on the right side and (x+4) on the left side.
5x (x+5) - 3(x+4)
(x+5)(x+4)(x+2) (x+5)(x+4)(x+2)
Focus on the top. 5x(x+5) will turn out to be 5x^2+25x. 3(x+4) will turn out to be 3x+12. Combine the two equations because now they are equal to each other and do the subtraction:
5x^2+25x - (3x+12) = 5x^2+22x-12 x cannot be -5, -4, -2
(x+5)(x+4)(x+2) (x+5)(x+4)(x+2)
Answer:
9 years old
Step-by-step explanation:
4x^2 + x + 3 = 0
x = [-1 +/- sqrt (1^2 - 4 * 4 * 3)] / 8 = - 1 +/- sqrt ( --47) / 8
= ( - 1 +/- sqrt47i) / 8 = -0.125 + 0.857i , -0.125 - 0.857i
Let's rewrite the binomial as:
Using the binomial expansion, we get:
For the 15th term, we want the term where r is equal to 14, because of the fact that the first term starts when r = 0. Thus, for the 15th term, we need to include the 0th or the first term of the binomial expansion.
Thus, the fifteenth term is:
