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shepuryov [24]
3 years ago
6

What are “like terms”? Why can we only add like terms?

Mathematics
1 answer:
SSSSS [86.1K]3 years ago
4 0

Answer:

Like terms are terms whose variables are the same.

We can only add them because if they does not have the same variable you do not actually know if the variables are equivalent and you will get the wrong answer.

Step-by-step explanation:

Ex: 4z,7z,108z,45z

Ex: 4w-2h+7a-2w

You can not add 4w and 7a because you do not know what the variables equal so you do not know the true number so add or subtract by.


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Answer:

x=23,y=15

Step-by-step explanation:

take away 8 from 38, divide it by 2, you get 15. add 8 to one of the set of two 15s. you get 23 and 15

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Factor the polynomial completely using the
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What are the excluded values of x for x+4/-3x^2+12x+36
LekaFEV [45]
I will assume that you meant:

(x+4)/(-3x^2+12x+36)  factor the denominator...

(x+4)/(-3(x^2-4x-12))

To factor a quadratic of the form ax^2+bx+c you need to find two values, j and k, which satisfy two conditions...

jk=ac=-12 and j+k=b=-4 so j and k must be 2 and -6 and the factors are then:

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Which equation represents a line which is perpendicular to the line 5x + 2y = 12?
o-na [289]

Answer:

<h2>          y = ²/₅ x - 3</h2>

Step-by-step explanation:

Changing to slope-intercept form:

5x + 2y = 12         {subtract 5x from both sides}

2y = -5x + 12         {divide both sides by 2}

y = -⁵/₂ x + 6

y=m₁x+b₁   ⊥   y=m₂x+b₂   ⇔    m₁×m₂ = -1

{Two lines are perpendicular if the product of theirs slopes is equal -1}

y =-⁵/₂ x + 1    ⇒     m₁ =  -⁵/₂

-⁵/₂× m₂ = -1      ⇒   m₂ = ²/₅

So, any line perpendicular to 5x + 2y = 12 must have slope m =²/₅

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