M+2 1/4=8 4/5 pls show your work . A) 6 3/1 B) 11 1/20 C) 6 11/20 D) 6 9/20

2 answers:

7 0

Please include the instructions. It appears that you want to find the value of m that makes m + 2 1/4 = 8 4/5, but you should state this explicitly.

Rewrite m + 2 1/4 = 8 4/5 as m + 2.25 = 8.80.
Subtr. 2.25 from both sides: m = 6.55 (answer)

This is also equal to 131/20 or 6 11/20.

Next time, would you please insert punctuation in your list of answers, to separate A from B, B from C, and so on.

nadya68 [22]3 years ago

4 0

There are several ways to do this. The best way to start is to make the denominators the same. The simplest common multiple to find is 20, since 4 × 5 is 20.

Multiply the numerator and denominator of 1/4 by 5, so it becomes 5/20.

Multiply the numerator and denominator of 4/5 by 4, so it becomes 16/20.

Notice that you didn't have to change the whole numbers.

Now subtract 2 5/20
(which was previously 2 1/4) from both sides of the equation.

On the left side you're left with just m.

On the right side you have 8 16/20 - 2 5/20.

You can subtract just the fractions first, 16/20 - 11/20, so you get 11/20. Subtracting the whole numbers, 8 - 2, gives you 6.

You then have
m = 6 11/20

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Find the midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9)

JulsSmile [24]

The midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9) is $\left(\frac{-9}{2}, \frac{-3}{2}\right)$

<u>Solution:</u>

Given, two points are (-6, 6) and (-3, -9)

We have to find the midpoint of the segment formed by the given points.

The midpoint of a segment formed by $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { and }\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$ is given by:

$\text { Mid point } \mathrm{m}=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

$\text { Here in our problem, } x_{1}=-6, y_{1}=6, x_{2}=-3 \text { and } y_{2}=-9$

Plugging in the values in formula, we get,

$\begin{array}{l}{m=\left(\frac{-6+(-3)}{2}, \frac{6+(-9)}{2}\right)=\left(\frac{-6-3}{2}, \frac{6-9}{2}\right)} \\\\ {=\left(\frac{-9}{2}, \frac{-3}{2}\right)}\end{array}$