Basically saying u can divide any number by 2 and 3 but not 4
Answer:
Step-by-step explanation:
Since the foci are at(0,±c) = (0,±63) and vertices (0,±a) = (0,±91), the major axis is the y- axis. So, we have the equation in the form (with center at the origin) 
.
We find the co-vertices b from b = ±√(a² - c²) where a = 91 and c = 63
b = ±√(a² - c²)
= ±√(91² - 63²)
= ±√(8281 - 3969)
= ±√4312
= ±14√22
So the equation is
<u> 12x³ - 12x </u><u /> = 3/7 - 1/56x² - 3/7 = 3/7 - 3/7 - 1/56x² = -1/56x²
28x³ - 56x² + 28x
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)
