dsillman2000

Because y = -5 and 5x-15y must also equal -5, we can calculate:

$5x-15(-5)=-5 \\ 5x = -80 \\ x=-16$

So your unknown, x = -16.

The greatest common factor (GCF) of both the numbers 100 and 14 is 2; if you divide the top and bottom by two you get:

$\frac{100}{14}\times\frac{1/2}{1/2}=\frac{50}{7}$

So 50/7 is your most accurate answer; The decimal approximation of this number is:

$\frac{50}{7}\approx7.143...$

One could simplify this number to scientific form, but I'm sure that it wouldn't help;

$1.70859375\times10^{8}$

$9c^2-(-3c^2-9c) \\\ \\ \text{Let's start by disbributing the negative:}\\\ \\ 9c^2+3c^2 +9c \\\ \\ \text{Now, you simply combine like terms.}\\ \text{This means, if the variable(}c\text{) has the same} \\ \text{power, then you can add those terms}\\ \text{together:}\\ \ \\ \underline{12c^2+9c}$

You might be in reference to "pi" which is defined as the number of diameter lengths you can fit around the circumference of a circle;

 $\pi=3.14159......$

Pi is a transcendental number, which means that it does not have a final decimal point, and cannot be expressed by a finite fraction. There are hundreds of ways to calculate pi via formulas, but I don't think you're as interested in that.

This question is the addition of two numbers in basic algebra;

$s=2\\ r=s+2\\ r=2+2=4$

Rick has 4 cookies.

Yeah my bad, I made a stupid mistake. It's y = 10. -80/-8 = 10.

First, distribute the 8 across what is in the parentheses, then subtract 56 from both sides, then divide both sides by -8 to get positive y:

$8(7-y) = -24\qquad\qquad\text{The original question}\\ 8\times 7 - 8\times y = -24\qquad\text{Distribute the 8 across the inside of the parentheses}\\ 56-8y=-24\qquad\qquad\quad\text{Simplify}\\ -8y=-80\qquad\qquad\qquad\text{Subtract 56 from both sides}\\ y=-10\qquad\qquad\qquad\qquad\text{Divide by -8 on both sides}$

So, your solution is y = -10

19. If you have 9 apples and add 10 more, you will have 19 apples:

$o\ o\ o\ o\ o\ o\ o\ o\ o\ o + o\ o\ o\ o\ o\ o\ o\ o\ o = \\ o\ o\ o\ o\ o\ o\ o\ o\ o\ o\ o\ o\ o\ o\ o\ o\ o\ o\ o$