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Lala15:
-10(2-6v)
ShadowOfDeath:
How we do this one is quite simple.
What we use is something called the distributive property.
For example, let's take the problem you have now.
We would "distribute" the -10 to both of the terms in the parenthesis.
Once you do that, the final product is a prime equation, so it's impossible to simplify. Therefore, your final answer is what you would get when you distribute -10 to the 2 and the 6v.
Isaacccc:
Idk🌚
jhonyy9:
@shadowofdeath wrote:
How we do this one is quite simple.
What we use is something called the distributive property.
For example, let's take the problem you have now.
We would "distribute" the -10 to both of the terms in the parenthesis.
Once you do that, the final product is a prime equation, so it's impossible to simplify. Therefore, your final answer is what you would get when you distribute -10 to the 2 and the 6v.
Attention !
-6v not 6v
ShadowOfDeath:
Oh. Thank you! I didn't notice that.
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Kaylee013:
the answer would be around 60v-20 if i did the math right
minionslover:
idk
bluebonnets:
60v-20
you would just distribute the -10 to the 2 and -6, and since a negative multiplied by a negative makes a positive then you would get 60
miazzzzlife:
in short the answer is 60v-20
Joe348:
In simpler words and explanations without giving a direct answer
You would distribute the -10 to both of the numbers 2 and -6 v
\[-10 \times 2 \]
\[-10 \times -6v\]
\[-10 \times 2 =-20\]
\[-10 \times -6v = 60v\]
Final answer :
\[60v-20\]