How do I factor y^2-2y-24

Mathematics

1 answer:

rusak2 [61]2 years ago

6 0

$y^2-2y-24=\\\\y^2+4y-6y-24=\\\\y(y+4)-6(y+4)=\\\\(y+4)(y-6)$

You might be interested in

An equilateral triangle has a perimeter of 15x + 3. What is the length of each side of the triangle?

maw [93]

An equilateral triangle have 3 congruent sides.

Divide the total (15x +3) by 3

(15x + 3)/3 = 5x + 1

Each side's length is 5x + 1

hope this helps

6 0

3 years ago

Calculate the following limit:

aleksklad [387]

$\lim_{x\to\infty}\dfrac{\sqrt x}{\sqrt{x+\sqrt{x+\sqrt x}}}=\\
\lim_{x\to\infty}\dfrac{\dfrac{\sqrt x}{\sqrt x}}{\dfrac{\sqrt{x+\sqrt{x+\sqrt x}}}{\sqrt x}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{\dfrac{x+\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{x}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\dfrac{\sqrt{x+\sqrt x}}{\sqrt{x^2}}}}=\\$
$\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{x+\sqrt x}{x^2}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{\sqrt x}{\sqrt{x^4}}}}}=\\\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{x}{x^4}}}}}=\\
\lim_{x\to\infty}\dfrac{1}{\sqrt{1+\sqrt{\dfrac{1}{x}+\sqrt{\dfrac{1}{x^3}}}}}=\\
=\dfrac{1}{\sqrt{1+\sqrt{0+\sqrt{0}}}}=\\$
$=\dfrac{1}{\sqrt{1+0}}=\\
=\dfrac{1}{\sqrt{1}}=\\
=\dfrac{1}{1}=\\
1
$

8 0

2 years ago

What is .7 of a mile

bearhunter [10]

Answer:

1232 yards or 3696 feet or 44352 inches or 0.7 of a mile

Step-by-step explanation:

4 0

2 years ago

QUESTION 1: Is the following equation proportional? How do you know? y=3x-4

nataly862011 [7]

Answer:

1: Yes because it is in the form y=mx+b

2: Yes because it is in the form y=kx

Step-by-step explanation:

I do this kind of stuff for math

7 0

2 years ago

I have a figure with the shaded region in the middle

stealth61 [152]

Answer:

Step-by-step explanation:

hmmmmmmmmmm

6 0

2 years ago