Answer:
it is 706.5 in sq
Step-by-step explanation:
Hey there!
Let's set up our expression:
(7a-6b+7)-(8a-2)
In order to simplify, we can use that subtraction sign and distribute it, using the distributive property. We have:
7a-6b+7-8a+2
Notice how it's plus two, because a negative times a negative two is a positive two. Now, it's a matter of finding the like terms and adding or subtracting them. These like terms can either have no variable, or have different coefficients but the same variable. That means our like terms are the 7a and -8a, and the 7 and 2. There's no like term for the 6b. That means we have:
(7a-8a) - 6b + (7+2) =
-a - 6b + 9
Hope this helps!
When ever you have percentages, it should be helpful to bear in mind you can express them as multipliers. In this case, it will be helpful.
So, if we let:
a = test score
b = target score
then, using the information given:
a = 1.1b + 1
a = 1.15b - 3
and we get simultaneous equations.
'1.1' and '1.15' are the multipliers that I got using the percentages. Multiplying a value by 1.1 is the equivalent of increasing the value by 10%. If you multiplied it by 0.1 (which is the same as dividing by 10), you would get just 10% of the value.
Back to the simultaneous equations, we can just solve them now:
There are a number of ways to do this but I will use my preferred method:
Rearrange to express in terms of b:
a = 1.1b + 1
then b = (a - 1)/1.1
a = 1.15b - 3
then b = (a + 3)/1.15
Since they are both equal to b, they are of the same value so we can set them equal to each other and solve for a:
(a - 1)/1.1 = (a + 3)/1.15
1.15 * (a - 1) = 1.1 * (a + 3)
1.15a - 1.15 = 1.1a + 3.3
0.05a = 4.45
a = 89
Answer:
112 $ is the original price
Step-by-step explanation:
you have to think that the original price multiplied by 0.75 (also known as 75%) gave the result of what he saves i.e 84 $. So we put up an equation with the original price multiplied by 75% which equals the 84 $ he saved on buying that item on sale
now, there are 12 months in a year, so 18 months is really 18/12 of a year, thus
![~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$4000\\ P=\textit{original amount deposited}\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ t=years\to \frac{18}{12}\dotfill &\frac{3}{2} \end{cases} \\\\\\ 4000=P[1+(0.05)(\frac{3}{2})]\implies 4000=P(1.075) \\\\\\ \cfrac{4000}{1.075}=P\implies 3720.93\approx P](https://tex.z-dn.net/?f=~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%281%2Brt%29%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5Cdotfill%20%26%20%5C%244000%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5C%5C%20r%3Drate%5Cto%205%5C%25%5Cto%20%5Cfrac%7B5%7D%7B100%7D%5Cdotfill%20%260.05%5C%5C%20t%3Dyears%5Cto%20%5Cfrac%7B18%7D%7B12%7D%5Cdotfill%20%26%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%204000%3DP%5B1%2B%280.05%29%28%5Cfrac%7B3%7D%7B2%7D%29%5D%5Cimplies%204000%3DP%281.075%29%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B4000%7D%7B1.075%7D%3DP%5Cimplies%203720.93%5Capprox%20P)
