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Masja [62]
2 years ago
5

Solve for X and Y thank u!

Mathematics
1 answer:
marusya05 [52]2 years ago
5 0

Answer:

I did most of the work 4 u :3

Step-by-step explanation:

Just do the last question

HOPE THIS HELPED :DDDDDDD

You might be interested in
 The area of a circle is less than 1 dm². Find the radius of the circle. The area of a circle is less than 1 dm². Find the rad
Ipatiy [6.2K]
\bf \textit{area of a circle}\\\\
A=\pi r^2\qquad 
\begin{cases}
\textit{let's say the Area is 1 exactly}\\
A=1
\end{cases}\implies 1=\pi r^2
\\\\\\
\cfrac{1}{\sqrt{\pi }}=r

if you drop the radius to less than that, then the Area will also drop to less than 1 unit, in this case it happens to be a decimeter
5 0
3 years ago
PLEASE HELP IS LINEAR INEQUALITIES !!
SashulF [63]

Answer:

THE ANSWER IS C

Step-by-step explanation:

BECAUSE BABABOOEEE

8 0
3 years ago
What is the common factor of the numerator and denominator in the expression <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%282x%2
mixer [17]

Answer:

Step-by-step explanation:

(x - 4)

What you are being asked to do is find the exact same binomial in the top as is in the bottom.

But there's a small catch. You must stipulate that x cannot equal 4. If it does, then you will get 0/0 which is undefined. You can't have that happening -- not at this level.

Any other value for x is fine.

5 0
3 years ago
Mary had five and one-half dollars. She spent tw wo and one-fourth dollars on a snack. How much money does Mary have left?
Levart [38]
5/1-1/4=4/3 i thinkkkkkkkkkk so
7 0
3 years ago
What does x+x+x+x+x-1+1+1+1= ?<br><br> (Giving this for 30 points)
77julia77 [94]

First, let's add all the x's.

x + x + x + x + x = 5x

5x - 1 + 1 + 1 + 1

Now, let's add together all the 1's that are being added (not the 1 that is being subtracted from 5x).


1 + 1 + 1 = 3

5x - 1 + 3

3 - 1 = 2

5x + 2

There's no way you can actually solve this unless you know the value of x; without x's value, this is as simplified as the expression is going to get.


Hope this helps!

7 0
3 years ago
Read 2 more answers
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