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3 years ago

What is 4/27 as an improper fraction?

Mathematics

2 answers:

Sever21 [200]3 years ago

6 0

That is too small to be an improper fraction.

Westkost [7]3 years ago

6 0

No, ThIs Equation is too small to be a proper fraction but you can simplify and reduce this

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Find the midpoint of the segment whose endpoints are (8,6) and (-2,-8).

inna [77]

Answer:

$\huge{ \fbox{ \sf{ \: ( \: 3 \: , \: - 1 \: )}}}$

Step-by-step explanation:

$\star{ \sf { \: Let \: the \: points \:be \: A \: and \: B}}$

$\star{ \sf{ \: Let \: \: A (8,6) \: be \: (x1 \:, y1) and \: B(-2,-8) \: be \: (x2 \:, y2)}}$

$\underline{ \sf{Finding \: the \: midpoint}}$

$\boxed{ \rm{Midpoint \: = \: ( \frac{x1 + x2}{2} \: , \: \frac{y1 + y2}{2} )}}$

$\mapsto{ \sf{Midpoint = ( \frac{8 + ( - 2)}{2} \: , \: \frac{6 + ( - 8)}{2} )}}$

$\mapsto { \sf{Midpoint = ( \frac{8 - 2}{2} \: , \: \frac{6 - 8}{2} }})$

$\mapsto{ \sf{Midpoint = ( \frac{6}{2} \: , \: \frac{ - 2}{2}) }}$

$\mapsto{ \sf{Midpoint = (3 \: \: , - 1)}}$

Hope I helped!

Best wishes!! :D

~ $\sf{TheAnimeGirl}$

5 0

3 years ago

Please help they’re 2 questions

IgorLugansk [536]

i think it’s is a for the first one

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4 years ago

Y=4x-6<br> 2x-3y=8<br> if this problem is to be solved by substitution , which is the correct step

Gekata [30.6K]

Answer:

x=1, y= -2

Step-by-step explanation:

Please see the attached pictures for full solution.

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Hopes this helps:

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