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3 years ago

What is solving systems of linear inequalities

2 answers:

6 0

It is when you have to find out if the line is linear or a parabola it is often used in the graphing portion of algebra but never really after that

erastova [34]3 years ago

6 0

The technique for solving these systems are pretty much simple. Here is an example. :) Solving systems of linear inequalities means graphing each individual inequality, and then finding the overlaps of various solutions.

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Can someone explain how I do this?

lorasvet [3.4K]

Answer:

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Step-by-step explanation:

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8 0

2 years ago

Read 2 more answers

Wally bakes muffins. in each batch of 12 muffins, he bakes 8 bluberry muffins write in simplest form the fraction of muffins tha

Flauer [41]

4/12 = 1/3 muffins are not blueberry

6 0

3 years ago

Here is my question:

sergiy2304 [10]

Answer:

$\frac{2}{3} - \frac{1}{2} = \frac{1}{6} \times \frac{8}{9} = \frac{4}{27}$

6 0

3 years ago

Prove the following statement using a proof by contraposition. Yr EQ,s ER, if s is irrational, then r + 1 is irrational.

sp2606 [1]

Answer:

I think that what you are trying to show is: If $s$ is irrational and $r$ is rational, then $r+s$ is rational. If so, a proof can be as follows:

Step-by-step explanation:

Suppose that $r+s$ is a rational number. Then $r$ and $r+s$ can be written as follows

$r=\frac{p_{1}}{q_{1}}, \,p_{1}\in \mathbb{Z}, q_{1}\in \mathbb{Z}, q_{1}\neq 0$

$r+s=\frac{p_{2}}{q_{2}}, \,p_{2}\in \mathbb{Z}, q_{2}\in \mathbb{Z}, q_{2}\neq 0$

Hence we have that

$r+s=\frac{p_{1}}{q_{1}}+s=\frac{p_{2}}{q_{2}}$

Then

$s=\frac{p_{2}}{q_{2}}-\frac{p_{1}}{q_{1}}=\frac{p_{2}q_{1}-p_{1}q_{2}}{q_{1}q_{2}}\in \mathbb{Q}$

This is a contradiction because we assumed that $s$ is an irrational number.

Then $r+s$ must be an irrational number.

3 0

3 years ago

Can y’all help find the surface area please and thank you

solong [7]

To find the area of a complex shape:

⇒ must cut the whole shape into a simpler shape

square with side lengths of 4 in
2 triangles with a base of 4 in. and a height of 8 in.
2 triangles with a base of 4 in. and a height of 10 in.

Let's calculate:

Area of 2 triangles with a base of 4 in. and a height of 8 in.

$Area = \frac{1}{2}*4*8 + \frac{1}{2}*4*8=16+16=32\\$

2 triangles with a base of 4 in. and a height of 10 in.

$Area = \frac{1}{2}*4*10 + \frac{1}{2}*4*10=20+20=40$

square with side lengths of 4 in

$Area = 4 * 4=16$

Total Area: 32 + 40 + 16 = 88 in²

Answer: 88 in²

Hope that helps!

8 0

1 year ago