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Alina [70]
3 years ago
9

How you solve x^2-xy+2x-2y

Mathematics
1 answer:
kap26 [50]3 years ago
7 0
You cant solve this because there is no matching variables like for example:

x^2-3xy+2x-2y-4+xy-5x+6y+8 and i solved this which the answer is x^2-3x+4y-2xy+2 because there are matching variables. 
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Help me answer these please will give brainlst
Naddika [18.5K]
I didn’t learn this yet but maybe
4 0
2 years ago
Estimate the square root of 150 to the nearest tenth.
muminat

the answer is 12.2 for 150 rounded to nearest tenth

5 0
3 years ago
Read 2 more answers
Suppose you have $100 in a savings account earning 2 percent interest a year. After five years, would you have more than $102, e
pentagon [3]
You would have more than $102.

Earning 2% a year in interest would get you $2 in interest the very first year:

100(0.02) = 2

Adding this to the amount in the account, you would have 100+2 = 102 after the first year.
6 0
3 years ago
Fond the equation of the line shown
zloy xaker [14]

Answer:

y+-1/2x-1

Step-by-step explanation:

The line is negative

It rises one digit and runs 2

and it crosses though -1 on the y axis

4 0
2 years ago
What’s the answer?and how do you get it
Kay [80]

the solid is made up of 2 regular octagons, 8 sides, joined up by 8 rectangles, one on each side towards the other octagonal face.

from the figure, we can see that the apothem is 5 for the octagons, and since each side is 3 cm long, the perimeter of one octagon is 3*8 = 24.

the standing up sides are simply rectangles of 8x3.

if we can just get the area of all those ten figures, and sum them up, that'd be the area of the solid.

\bf \textit{area of a regular polygon}\\\\ A=\cfrac{1}{2}ap~~ \begin{cases} a=apothem\\ p=perimeter\\[-0.5em] \hrulefill\\ a=5\\ p=24 \end{cases}\implies A=\cfrac{1}{2}(5)(24)\implies \stackrel{\textit{just for one octagon}}{A=60} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{two octagon's area}}{2(60)}~~+~~\stackrel{\textit{eight rectangle's area}}{8(3\cdot 8)}\implies 120+192\implies 312

7 0
3 years ago
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