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vladimir2022 [97]
3 years ago
13

6(4a^2+11)- 13(3a-2a^2)-28 please help quick!!

Mathematics
2 answers:
attashe74 [19]3 years ago
5 0
Split it out and solve it !!

6(4a^2+11)- 13(3a-2a^2)-28

24a² + 66 - 39a +26a² - 28

50a² - 39a + 38
Sergio [31]3 years ago
4 0

(24a^2+66)- (39a-26a^2)-28

24a^2 + 66 - 39a + 26a^2 - 28

Combine like terms:

50a^2 + 38 - 39a

Divide the equation by 2:

25a + 19 - 19.5a

44.5a + 19 Is the simplest form of this equation.


I hope this helps! :)


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Read 2 more answers
The sum of three consecutive integers is 114. what are the three integers?
RSB [31]

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3 consecutive integers can be represented as followed.

X ⇒ <em>first integer</em>

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<em />

Since the sum of our 3 consecutive integers is 114, we can set up an equation to represent this.

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÷3    ÷3   ← <em>divide both sides of the equation by 3</em>

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X + 1 ⇒ <em>second integer   = 38</em>

X + 2 ⇒ <em>third integer   = 39</em>

<em />

<em>Therefore, our 3 consecutive integers are 37, 38, and 39.</em>

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8 0
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