Mathematics, 30.06.2019 09:30 Jimm6500

What’s the least common multiple of x^2-2x-15 and x^2+2x-3?

Answers

$\large\boxed{LCM(x^2-2x-15,\ x^2+2x-3)=(x+3)(x-5)(x-1)}\\\boxed{=x^3-3x^2-13x+15}$

Step-by-step explanation:

$x^2-2x-15=x^2+3x-5x-15=x(x+3)-5(x+3)=(x+3)(x-5)\\\\\\x^2+2x-3=x^2+3x-1x-3=x(x+3)-1(x+3)=(x+3)(x-1)\\\\LCM(x^2-2x-15,\ x^2+2x-3)=(x+3)(x-5)(x-1)\\\\=(x^2-2x-15)(x-1)\qquad\text{use FOIL}\\\\=(x^2)(x)+(x^2)(-1)+(-2x)(x)+(-2x)(-1)+(-15)(x)+(-15)(-1)\\\\=x^3-x^2-2x^2+2x-15x+15\qquad\text{combine like terms}\\\\=x^3+(-x^2-2x^2)+(2x-15x)+15\\\\=x^3-3x^2-13x+15$

Answer from: Quest

p ÷ r = tan(x)

step-by-step explanation:

given that there is a figure below shows a right triangle: a right triangle is shown with hypotenuse equal to q units and the length of the legs equal to p units and r units.

graph is missing so i will create the graph as per given information.

the angle between the legs having lengths p units and q units is y degrees. the angle between the legs having lengths r units and q units is x degrees.

now we need to find about what is p ÷ r equal to.

so we can use trigonometric ratio tan(x)=[altitude]/[base] to find the value of p ÷ r, because p is the altitude and r is the base.

$\tan\left(x\right)=\frac{\left[altitude\right]}{\left[base\right]}=\frac{p}{r}$