$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{3}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{-6}~,~\stackrel{y_2}{-6}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-6+3}{2}~~,~~\cfrac{-6+5}{2} \right)\implies \left(\cfrac{-3}{2}~~,~~\cfrac{-1}{2} \right)\implies \left( -1\frac{1}{2}~,~-\frac{1}{2} \right)$

3 0

2 years ago

Which of the following statements says that a number is between -3 and 3?

Shalnov [3]

So, if the <span>number is between -3 and 3, It couldn't be more than 3, so C isn't the answer. It also doesn't have to be equal to 3.

B, </span><span>|x| < 3, would be your answer.
Hope this helps!</span>

4 0

2 years ago

Read 2 more answers

I HELP NEED ASAP PLZ HELP ME

Artemon [7]

Step-by-step explanation:

hope it helps.........

6 0

2 years ago

Read 2 more answers

What is the solution to the system of equations? Use the substitution method. {y=−2x+53x−4y=2

klio [65]

$y = - 2x + 53x - 4y = 2 \\ y = 51x - 4y = 2 \\ > y = 51 x - 4y \\ > 51x - 4y = 2$
$y = 51x - 4y \\ 5y = 51x \\ y = \frac{51}{5} x$
Substitute 51/5x into
$51x - 4( \frac{51}{5} x) = 2 \\ 51x - 40.8x = 2 \\ 10.2x = 2 \\ x = 2 \times \frac{10}{102} \\ x = \frac{10}{51} \: aka \: 0.196$
Substitute that into
$y = \frac{51}{5} \times \frac{10}{51} \\ y = 2$
Hope this helps. Answer should be correct. Unless your quadratic equation is wrongly stated.

8 0

2 years ago