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victus00 [196]
3 years ago
14

I need to get rid of the parenthesis and match like terms of -5(2x-3y)+2(x+2y)

Mathematics
2 answers:
irga5000 [103]3 years ago
5 0

Answer:

-8x + 19y

Step-by-step explanation:

distribute the -5

-10x + 15y + 2(x + 2y)

distribute the 2

-10x +15y + 2x + 4y

arange  the terms

-10x + 2x + 15y + 4y

combine

-8x + 19y

kakasveta [241]3 years ago
4 0

Answer:

-8x+19y

Step-by-step explanation:

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The parallelogram ABCD is translated 7 units to the left to create A'B'C'D'.

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5 0
3 years ago
Table of value y = -4x - 12
Scorpion4ik [409]
X Y
————-
-2. -4
-1. -8
0. -12
1. -16
2. -20
8 0
2 years ago
Overline MD cong overline LS additional information is necessary to show that triangle MTD cong triangle LGS by SSS?
aivan3 [116]

Answer:

TD \cong GS

Step-by-step explanation:

See comment for complete question

Given:

TM \cong GL

MD \cong LS

Required

The information that shows \triangle MTD \cong \triangle LGS by SSS

By SSS implies that, the three sides of both triangles are congruent

Already, we have:

TM \cong GL

MD \cong LS

The third side of \triangle MTD is TD

The third side of \triangle LGS is GS

So, for both to be congruent by SSS, the third sides must be congruent

i.e.

TD \cong GS

4 0
2 years ago
Add the following rational number =<br> 3/11 and -5/12
tatiyna

Answer:

-19/132

Step-by-step explanation:

To add fractions, find the lowest common multiple of the two denominators and multiply accordingly to make both denominators the same. In this case, the lowest common multiple of 11 and 12 is 132, so we need to multiply the first number by 12 and the second by 11. So we get \frac{36}{132} + (-\frac{55}{132} ) = \frac{36-55}{132}=\frac{-19}{32}-\frac{19}{132}

4 0
2 years ago
If (ax^2 + 3x + 2b) - (5x^2+bx-3c)=3x^2-9, what is the value of A + B + C?
Rina8888 [55]

Answer:

We have the equation:

(ax^2 + 3x + 2b) - (5x^2+bx-3c)=  3x^2 - 9

First, move all to the left side.

(ax^2 + 3x + 2b) - (5x^2+bx-3c) - 3x^2 + 9 = 0

Now let's group togheter terms with the same power of x.

(a - 5 - 3)*x^2 + (3 - b)*x + (2b + 3c + 9) = 0.

This must be zero for all the values of x, then the things inside each parenthesis must be zero.

1)

a - 5 - 3 = 0

a = 3 + 5 = 8.

2)

3 - b = 0

b = 3.

3)

2b + 3c + 9 = 0

2*3 + 3c + 9 = 0

3c = -6 - 9 = -15

c = -15/3 = -5

Then we have:

a = 8, b = 3, c = -5

a + b + c = 8 + 3 - 5 = 6

8 0
3 years ago
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