All categories

3 years ago

What is a solution to a system of inequality's?

Mathematics

2 answers:

Damm [24]3 years ago

8 0

Make all equal or try to even it out depending on what the system is about

Reptile [31]3 years ago

8 0

Answer:

You can graph a linear inequality in two variables such as x or y, first get y solo on one side. The solution of the structure of inequalities is the intersection region of all the solutions in the system.

Step-by-step explanation:

You might be interested in

It doesn’t have the plus sign

3241004551 [841]

+

there u go if that helps u can copy and paste it

5 0

2 years ago

Read 2 more answers

Raquel works in a pet store every day after school and sometimes on weekends. She earns $5.20 per hour. If Raquel works 21 hours

riadik2000 [5.3K]

Answer:

Step-by-step explanation:

$5.20 is about $5.00, 21 is about 20, so $5.00 * 20 *4 = $400.

Checking this: $5.20 * 21 * 4 = $436.80

8 0

2 years ago

Read 2 more answers

I need to show my work, could anyone help? (need by tonight 3/10/20

marishachu [46]

Answer:

14 gallons....$42

Step-by-step explanation:

The car requires one gallon every 32 miles...

The car requires c gallons for 448 miles

c = 448 × 1 / 32

= 14 gallons..

One gallon = $3

Fourteen gallons = ??

14 × 3 / 1 = $42

3 0

3 years ago

Please answer correctly !!!!!!!!! Will<br> Mark brainliest !!!!!!!!!!!!!

il63 [147K]

Answer:

$x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}$

Step-by-step explanation:

$w=-5\left(x-8\right)\left(x+4\right)\\\mathrm{Expand\:}-5\left(x-8\right)\left(x+4\right):\quad -5x^2+20x+160\\w=-5x^2+20x+160\\Switch\:sides\\-5x^2+20x+160=w\\\mathrm{Subtract\:}w\mathrm{\:from\:both\:sides}\\-5x^2+20x+160-w=w-w\\Simplify\\-5x^2+20x+160-w=0\\Solve\:with\:the\:quadratic\:formula\\\mathrm{Quadratic\:Equation\:Formula:}\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=-5,\:b=20,\:c=160-w:\quad x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}\\x=\frac{-20+\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20+\sqrt{-20w+3600}}{10}\\x=\frac{-20-\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20-\sqrt{-20w+3600}}{10}\\The\:solutions\:to\:the\:quadratic\:equation\:are\\x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}$

6 0

3 years ago

Need help on this, what is the correct name for this number?

Flauer [41]

Answer:

I believe its the third one

6 0

3 years ago

Read 2 more answers