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VLD [36.1K]
2 years ago
9

Ayo i need help this is vv hard or i’m jus dumb.

Mathematics
1 answer:
erastovalidia [21]2 years ago
4 0
It goes straight up I hope this helps if not in sorry
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Will give brainliest!!
Inessa05 [86]
It’s actually B Bc the little 1 by the 10 is meaning that you are gonna multiply 10x10 one time. Then multiply by 7. 10x10=100 100x7=700 so there for your answer is B
4 0
3 years ago
(5x2+14x-15)-(3x2+5x+7) simplify
sleet_krkn [62]

Answer:

-241

Step-by-step explanation:

Hope this helps

8 0
2 years ago
Read 2 more answers
Rewrite the equation below so that it does not have fractions.
rjkz [21]
.75x+4=.75

hope this helps!
4 0
3 years ago
Read 2 more answers
I really don't get it can you give me the answer then explain why it is that one please ​
Ronch [10]

Answer:

\frac{100}{9} is the answer.

Step-by-step explanation:

Well the first thing we would do is simplify all the exponents first inside the parenthesis. So 2^-3 is equal to 1/8. To solve negative exponents you would just divide the number instead of multiply, so 2 divided by 2 divided by 2 is equal to ⅛ or .125

Next would be to do the 2^-9. This would equal 1/512

Now would be to simplify 5^0. Anything to the power of 0 is always one. So now we simplify all the products meaning 7*5*2 and 7*3.  

Now we would simplify both fractions.

The picture are pretty much self explanatory. If you have any questions feel free to ask in the comments - Mark

Also when you have the chance please mark me brainliest.

7 0
3 years ago
Read 2 more answers
Sigma notation from sum. pls help?? sorry for the calc spam :P
givi [52]
To the risk of sounding redundant,

\bf \begin{array}{ccccccccc}
k=2&&k=3&&k=4&&k=5&&k=6\\\\
\frac{\underline{2}+2}{2(\underline{2})}&&\frac{\underline{3}+2}{2(\underline{3})}&&\frac{\underline{4}+2}{2(\underline{4})}&&\frac{\underline{5}+2}{2(\underline{5})}&&\frac{\underline{6}+2}{2(\underline{6})}\\\\
1&+&\cfrac{5}{6}&+&\cfrac{6}{8}&+&\cfrac{7}{10}&+&\cfrac{8}{12}
\end{array}

and surely you'd know what that is.
7 0
3 years ago
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