Answer:
Whenever
.
Step-by-step explanation:
We can play around with some numbers and develop some rules for this equation.
Note that the number 5 and -5 are used here, so let's try using 5 as a.
![|5-5| = 5-5\\|0| = 0\\0 = 0](https://tex.z-dn.net/?f=%7C5-5%7C%20%3D%205-5%5C%5C%7C0%7C%20%3D%200%5C%5C0%20%3D%200)
So 5 works. Let's try a random number like 3.
![|3-5| = 5-3\\|-2| = 2\\2 = 2](https://tex.z-dn.net/?f=%7C3-5%7C%20%3D%205-3%5C%5C%7C-2%7C%20%3D%202%5C%5C2%20%3D%202)
Okay, with this info we know that we might be able to develop one rule that
. Just to test, let's try 0, -3, and -5.
![|0-5| = 5-0\\|-5| = 5\\5 = 5](https://tex.z-dn.net/?f=%7C0-5%7C%20%3D%205-0%5C%5C%7C-5%7C%20%3D%205%5C%5C5%20%3D%205)
Zero works.
![|-3 - 5| = 5-(-3)\\|-8| = 8\\8 = 8](https://tex.z-dn.net/?f=%7C-3%20-%205%7C%20%3D%205-%28-3%29%5C%5C%7C-8%7C%20%3D%208%5C%5C8%20%3D%208)
-3 works.
![|-5 -5| = 5-(-5)\\|-10| = 10\\10 = 10](https://tex.z-dn.net/?f=%7C-5%20-5%7C%20%3D%205-%28-5%29%5C%5C%7C-10%7C%20%3D%2010%5C%5C10%20%3D%2010)
-5 works. Now, this might stop here making the equation
, so let's test a number outside of -5 - say -20.
![|-20 - 5| = 5-(-20)\\|-25| = 25\\25 = 25](https://tex.z-dn.net/?f=%7C-20%20-%205%7C%20%3D%205-%28-20%29%5C%5C%7C-25%7C%20%3D%2025%5C%5C25%20%3D%2025)
Yes! This works, so a works for this equation as long as
.
Hope this helped!
Answer:
25
Step-by-step explanation:
(-5)^2+10(-5)(2)+25(2)^2
=25+(-100)+100
=25
liters is the amount to be drained out and replaced
<em><u>Solution:</u></em>
40 % antifreeze solution in 16 liter radiator
Let "x" be the amount drained from radiation and replaced with pure antifreeze
To obtain a 60 % antifreeze solution
The original solution is 16 liter, 40% of which is antifreeze
You want the solution to be 60% antifreeze:
60 % x 16 = ![\frac{60}{100} \times 16 = 9.6](https://tex.z-dn.net/?f=%5Cfrac%7B60%7D%7B100%7D%20%5Ctimes%2016%20%3D%209.6)
You will remove x liters of the 40% solution and replace it with x liters pure (100%) antifreeze.
![40 \% (16 - x) + 100 \% \times x = 60 \% \times 16](https://tex.z-dn.net/?f=40%20%5C%25%20%2816%20-%20x%29%20%2B%20100%20%5C%25%20%5Ctimes%20x%20%3D%2060%20%5C%25%20%5Ctimes%2016)
Let us solve expression for "x"
![\frac{40}{100} \times (16 - x) + \frac{100}{100} \times x = \frac{60}{100} \times 16\\\\0.4(16-x) + x = 0.6 \times 16\\\\6.4 - 0.4x + x = 9.6\\\\6.4 + 0.6x = 9.6\\\\0.6x = 3.2\\\\x = 5.33\\\\x = 5\frac{1}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B40%7D%7B100%7D%20%5Ctimes%20%2816%20-%20x%29%20%2B%20%5Cfrac%7B100%7D%7B100%7D%20%5Ctimes%20x%20%3D%20%5Cfrac%7B60%7D%7B100%7D%20%5Ctimes%2016%5C%5C%5C%5C0.4%2816-x%29%20%2B%20x%20%3D%200.6%20%5Ctimes%2016%5C%5C%5C%5C6.4%20-%200.4x%20%2B%20x%20%3D%209.6%5C%5C%5C%5C6.4%20%2B%200.6x%20%3D%209.6%5C%5C%5C%5C0.6x%20%3D%203.2%5C%5C%5C%5Cx%20%3D%205.33%5C%5C%5C%5Cx%20%3D%205%5Cfrac%7B1%7D%7B3%7D)
Thus
liters is the amount to be drained out and replaced
40,400 = 40,000(1 + 0.06/6)^(6t)
n = 0.1667
The money has been invested for 2 months.
Answer: -11
Step-by-step explanation:
4(-2)-3
-8-3
=-11