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ddd [48]
3 years ago
5

Combine like terms: ab-a2+42-5ab+3a2+10

Mathematics
1 answer:
KATRIN_1 [288]3 years ago
4 0

Answer:

   -4AB+2A2+52

Step-by-step explanation:

1. FIRST WE PUT THE LIKE TERMS TOGETHER (REARRANGE THE PROBLEM)

AB-5AB-A2+3A2+42+10

2. SIMPLIFY ALL LIKE TERMS

-4AB+2A2+52

YOUR DONE!!!!!!

HOPE I HELPED

CAN I GET BRAINLIEST PLS? (DESPERATELY TRYING TO LEVEL UP)

         -ZYLYNN JADE

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Pls help im really bad at math no links pls​
lawyer [7]

Answer:

-\frac{1}{2}

Step-by-step explanation:

-\frac{5}{6} +\frac{6}{18}

-\frac{5}{6} +\frac{2}{6}        reduced

\frac{-5+2}{6}            transformed the expression to make it easier

-\frac{3}{6}               add

-\frac{1}{2}               reduce

8 0
2 years ago
A credit report maintains negative financial information for how many years?
Dimas [21]
I believe the answer is B
8 0
3 years ago
Please help! With steps included :)
Len [333]

Answer:

Top question: -7°

Bottom question : 18°

Step-by-step explanation:

Top question:

The base angles of am isosceles triangle are congruent:

x + 46 = 39 because the base angles of an isosceles triangle are congruent

x + 46° = 39°

x = -7°

Bottom question

The missing angle is 3x because the base angles are congruent

To find the value of x, we must add all the angles inside the triangle, and find the value of x that makes it equal to 180°

3x + 3x + 4x = 180°

10x = 180°

x = 18°

Sorry for the late post!

-Chetan K

8 0
3 years ago
Can someone help me solve this step by step?
patriot [66]

Answer:

41 and 77

Step-by-step explanation:

x: his age the first time he went to space

y: his age the second time he went to space

the second time he went to space he was 5 years younger than twice the age when he went his first time, then:

y = 2x - 5

the sum of both ages is 118, then:

x + y = 118

If you replace y = 2x - 5 in x + y =118:

x + 2x - 5 = 118

3x = 123

x = 41

replacing x:

41 + y = 118

y = 118 - 41

y = 77

so when he was 77, he was 5 years younger than twice 41 ( 41*2 = 82 => 82 - 5 =77)

4 0
3 years ago
Solve simultaneously <br> Y=x-2<br> Y^2=x
omeli [17]
\left \{ {{y=x-2} \atop {\ y^2=x}} \right. \\ \\ \\&#10;y=x-2 \\&#10;y^2=(x-2)^2=x^2-4x+4 \\ \\ \\&#10;y^2=x \\&#10;x^2-4x+4=x \\&#10;x^2-5x+4=0


a=1 \\ b=-5 \\ c=4 \\ \\ x=\frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}=\frac{5 \pm \sqrt{25 - 16}}{2}=\frac{5 \pm \sqrt{9}}{2}=\frac{5 \pm 3}{2} \\ \\&#10;x=\frac{5+3}{2} \ \lor \ x=\frac{5-3}{2} \\ \\&#10;x=\frac{8}{2} \ \lor \ x=\frac{2}{2} \\ \\&#10;x=4 \ \lor \ x=1


y=4-2 \ \lor \ y=1-2 \\&#10;y=2 \ \lor \ y=-1 \\ \\ \\&#10; \left \{ {{x=4} \atop {y=2}} \right.  \ \lor \  \left \{ {{x=1} \atop {y=-1}} \right.
8 0
3 years ago
Read 2 more answers
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