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3 years ago

9

What is the value of :-=》

la1" title=" \frac{4 - 4}{5 - 5} " alt=" \frac{4 - 4}{5 - 5} " align="absmiddle" class="latex-formula">

1 answer:

Tresset [83]3 years ago

4 0

Answer:

the value is undefined

Step-by-step explanation:

hey there,

->

$\frac{4 - 4}{5 - 5}$

->

$\frac{0}{0}$

0 divided by 0 is undefined

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The diagonals bisect each other.

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The diagonals of a parallelogram bisect each other.

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Plz help if you can<br> Thank ya

koban [17]

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$90\pi$

Step-by-step explanation:

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3 years ago

What is the solution set for 2x+5y>-1 and 4x-3<-3?

bekas [8.4K]

PROBLEM ONE

•

Solving for x in 2x + 5y > -1.

•

Step 1 ) Subtract 5y from both sides.

2x + 5y > -1

2x + 5y - 5y > -1 - 5y

2x > -1 - 5y

Step 2 ) Divide both sides by 2.

2x > -1 - 5y

$\displaystyle\frac{2x}{2} > \displaystyle\frac{-1 - 5y}{2}$

$\displaystyle\ x > \frac{-1 - 5y}{2}$

So, the solution for x in 2x + 5y > -1 is...

$\displaystyle\ x > \frac{-1 - 5y}{2}$

•

Solving for y in 2x + 5y > -1.

•

Step 1 ) Subtract 2x from both sides.

2x + 5y > -1

2x - 2x + 5y > -1 - 2x

5y > -1 - 1x

Step 2 ) Divide both sides by 5.

5y > -1 - 1x

$\displaystyle\frac{5x}{5} > \frac{-1 -1x}{5}$

$\displaystyle\ x > \frac{-1 -1x}{5}$

So, the solution for y in 2x + 5y > -1 is...

$\displaystyle\ x > \frac{-1 -1x}{5}$

•

PROBLEM TWO

•

Solving for x in 4x - 3 < -3.

•

Step 1 ) Subtract 3 from both sides.

4x - 3 < -3

4x -3 - 3 < -3 - 3

4x < 0

Step 2 ) Divide both sides by x.

4x < 0

$\displaystyle\frac{4x}{4}$

x < 0

So, the solution for x in 4x - 3 < -3 is...

x < 0

•

•

- <em>Marlon Nunez</em>

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