I need help with 8 i don't understand it i get. idk how to figure out what jm is

Mathematics

1 answer:

Misha Larkins [42]3 years ago

8 0

The segment bisector of line JK would be L since it passes through M
How to find JM is that you would have to solve 3x+15=8x+25 to get x and once you find x you just plug it into the equation for JM, 3x+15
3x+15=8x+25 (subtract 8x on both sides)
-5x+15=25 (subtract 15 on both sides)
-5x=10 (Divide on both sides)
x=-2
Plug that in:
3(-2)+15
-6+15
JM=9

You might be interested in

The following triangles are similar. State the postulate or theorem that proves they are similar along with the correct scale fa

Nadusha1986 [10]

I would have to say SAS because they give 2 sides and an angle. Also 3:4 because the one side is 3 and add 1 for 4, same for 30 and 40.

6 0

2 years ago

How do you solve 32 - 6x = 53?

katrin2010 [14]

$\large\boxed{x=-\frac{7}{2}}$

To solve for $x$ , we need to isolate it on one side of the equation.

The most important part of this is knowing that whatever we do to one side of the equation, we must also do to the other.

Subtract 32 from both sides of the equation.

$\begin{aligned}32-32-6x&=53-32\\-6x&=21\end{aligned}$

Divide both sides of the equation by $-6$ .

$\begin{aligned}\frac{-6x}{-6}&=\frac{21}{-6}\\x&=\boxed{-\frac{7}{2}}\end{aligned}$

7 0

1 year ago

Read 2 more answers

What is the completely simplified equivalent of 2/(5+i)?

Bezzdna [24]

namely, let's rationalize the denominator in the fraction, for which case we'll be using the conjugate of that denominator, so we'll multiply top and bottom by its conjugate.

so the denominator is 5 + i, simply enough, its conjugate is just 5 - i, recall that same/same = 1, thus (5-i)/(5-i) = 1, and any expression multiplied by 1 is just itself, so we're not really changing the fraction per se.

$\bf \cfrac{2}{5+i}\cdot \cfrac{5-i}{5-i}\implies \cfrac{2(5-i)}{\stackrel{\textit{difference of squares}}{(5+i)(5-i)}}\implies \cfrac{2(5-i)}{\stackrel{\textit{recall }i^2=-1}{5^2-i^2}}\implies \cfrac{2(5-i)}{25-(-1)} \\\\\\ \cfrac{2(5-i)}{25+1}\implies \cfrac{2(5-i)}{26}\implies \cfrac{5-i}{13}$