now that we know what are the x-values, what are the y-values? well, we can just use the 2nd equation, since we know that y = x - 28, then
![\bf y = x - 28\implies \stackrel{\textit{when x = 16}}{y = 16 - 28}\implies y = -12 \\\\\\ y = x - 28\implies \stackrel{\textit{when x = 12}}{y = 12 - 28}\implies y = -16 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\stackrel{x}{16}~~,~~\stackrel{y}{-12})\qquad,\qquad (\stackrel{x}{12}~~,~~\stackrel{y}{-16})~\hfill](https://tex.z-dn.net/?f=%5Cbf%20y%20%3D%20x%20-%2028%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bwhen%20x%20%3D%2016%7D%7D%7By%20%3D%2016%20-%2028%7D%5Cimplies%20y%20%3D%20-12%20%5C%5C%5C%5C%5C%5C%20y%20%3D%20x%20-%2028%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bwhen%20x%20%3D%2012%7D%7D%7By%20%3D%2012%20-%2028%7D%5Cimplies%20y%20%3D%20-16%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%28%5Cstackrel%7Bx%7D%7B16%7D~~%2C~~%5Cstackrel%7By%7D%7B-12%7D%29%5Cqquad%2C%5Cqquad%20%28%5Cstackrel%7Bx%7D%7B12%7D~~%2C~~%5Cstackrel%7By%7D%7B-16%7D%29~%5Chfill)
<span>x^2 + 8x – 48 = 0
That is a quadratic equation where
a = 1
b = 8
c = -48
We can solve it by using the quadratic formula:
x = [-b +-sq root (b^2 - 4ac)] / 2a
x = [-8 +- sq root (64 -4*1*-48)] / 2*1
</span><span>x = [-8 +- sq root (256)] / 2
x = [-8 +-16] / 2
x1 = 8 / 2 = 4
x2 = -24 / 2 = -12
</span>
Answer:
24 pencils : 64 pens
Step-by-step explanation:
pencils : pens
12:32
Twice as many pencils mean multiply each by 2
12*2 : 32*2
24: 64
24 pencils : 64 pens
Answer:
A = 1/2
Step-by-step explanation:
"A+6=9-5A" is already an equation, and your task here is to solve it for A.
Adding 5A to both sides results in
A + 5A + 6 = 9, or
6A + 6 = 9.
Subtracting 6 from both sides results in
6A = 3.
Thus, A = 3/6, or, in simplest terms, A = 1/2.
