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Zanzabum
3 years ago
8

How would you solve (3a-4)^2?

Mathematics
2 answers:
Nesterboy [21]3 years ago
6 0
Well (3a - 4)^2 is equivalent to (3a - 4) * (3a - 4)
How I would solve that is by firstly multiplying (3a - 4) by 3a (first part of second bracket) then adding that to (3a - 4) * - 4 (second part of second bracket)

That would equal
(9a^2 - 12a) + (-12a + 16)
= 9a^2 - 12a - 12a + 16
= 9a^2 - 24a +16

Sorry if my answer is rubbish
Vinvika [58]3 years ago
3 0
(3a - 4)^2 =
(3a - 4)(3a - 4) =
3a(3a - 4) - 4(3a - 4) =
9a^2 - 12a - 12a + 16 =
9a^2 - 24a + 16 <== I am not getting what ur teacher got

is the problem written correctly ?
Because (3a - 3)^2 = 9a^2 - 18a + 9
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3 years ago
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Can someone tell me if this is right?
Luda [366]

Answer: (no promises this is exactly right)

-2r+6

Step-by-step explanation:

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3 years ago
For the right triangle shown, the lengths of two sides are given. Find the third side. Leave your answer in simplified, radical
sleet_krkn [62]

Answer: c=4\sqrt{2}

Step-by-step explanation:

For this exercise you need to apply the Pythagorean Theorem. This is:

h^2=l^2+m^2

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In this case, you can identify that:

h=c\\\\l=a=4\\\\m=b=4

Knowing these values, you can substitute them into  h^2=l^2+m^2:

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Now you must solve for "c":

c=\sqrt{4^2+4^2}

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To simplify the result:

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6 0
3 years ago
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hram777 [196]

Answer:

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1


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24/2

12/2

6/2

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GCF = 2³ = 8

8 0
2 years ago
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