Answer:
Patricia spent $24 on presents, while Donald spent $12 and Carl spent $36.
Step-by-step explanation:
The problem is to solve how much money each person spent on presents in the year, through a series of equivalences. In order to determine how much money each one spent, the following equation must be proposed:
Patricia = X
Donald = 1 / 2X
Carl = 3/2 X
X + 1 / 2X + 3 / 2X = 72
1X + 0.5 X + 1.5 X = 72
3X = 72
X = 72/3
X = 24
Thus, Patricia spent $ 24 on presents, while Donald spent $ 12 (2/24) and Carl spent $ 36 (1.5 x 24).
Set up a system of equations.
d + q = 75
.1d + .25q = 14.25
Multiply the first equation by -.1 so that the "d's" will cancel.
-.1d -.1q = -7.5
.1d +.25q = 14.25
Now add the two equations together
.15q = 6.75
divide both sides by .15
q = 45
There are 45 quarters in the jar
Answer:
Sodas: 160
Hot Dogs: 80
Step-by-step explanation:
Sodas: s
Hot Dogs: d
s+d=240
s=2d
3d=240
d=80
s=160
if we have a number like say hmm 4, and we say hmmm √4 is ±2, it simply means, that if we multiply that number twice by itself, we get what's inside the root, we get the 4, so (+2)(+2) = 4, and (-2)(-2) = 4, recall that <u>minus times minus = plus</u>.
so, any when we're referring to even roots like ![\bf \sqrt[2]{~~},\sqrt[4]{~~},\sqrt[6]{~~}....](https://tex.z-dn.net/?f=%5Cbf%20%5Csqrt%5B2%5D%7B~~%7D%2C%5Csqrt%5B4%5D%7B~~%7D%2C%5Csqrt%5B6%5D%7B~~%7D....)
, the positive number, that can multiply itself an even amount of times, will produce a valid value, BUT the negative number that multiply itself an even amount of times, will also produce a valid value.
now, that's is not true for odd roots like ![\bf \sqrt[3]{~~},\sqrt[5]{~~},\sqrt[7]{~~}....](https://tex.z-dn.net/?f=%5Cbf%20%5Csqrt%5B3%5D%7B~~%7D%2C%5Csqrt%5B5%5D%7B~~%7D%2C%5Csqrt%5B7%5D%7B~~%7D....)
, because the multiplication of the negative number will not produce a valid value, let's put two examples on that.
![\bf \sqrt[3]{27}\implies \sqrt[3]{3^3}\implies 3\qquad because\qquad (3)(3)(3)=27 \\\\\\ however\qquad (-3)(-3)(-3)\ne 27~\hspace{8em}(-3)(-3)(-3)=-27 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \sqrt[3]{-125}\implies \sqrt[3]{-5^3}\implies -5\qquad because\qquad (-5)(-5)(-5)=-125 \\\\\\ however\qquad (5)(5)(5)\ne -125~\hspace{10em}(5)(5)(5)=125](https://tex.z-dn.net/?f=%5Cbf%20%5Csqrt%5B3%5D%7B27%7D%5Cimplies%20%5Csqrt%5B3%5D%7B3%5E3%7D%5Cimplies%203%5Cqquad%20because%5Cqquad%20%283%29%283%29%283%29%3D27%0A%5C%5C%5C%5C%5C%5C%0Ahowever%5Cqquad%20%28-3%29%28-3%29%28-3%29%5Cne%2027~%5Chspace%7B8em%7D%28-3%29%28-3%29%28-3%29%3D-27%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Csqrt%5B3%5D%7B-125%7D%5Cimplies%20%5Csqrt%5B3%5D%7B-5%5E3%7D%5Cimplies%20-5%5Cqquad%20because%5Cqquad%20%28-5%29%28-5%29%28-5%29%3D-125%0A%5C%5C%5C%5C%5C%5C%0Ahowever%5Cqquad%20%285%29%285%29%285%29%5Cne%20-125~%5Chspace%7B10em%7D%285%29%285%29%285%29%3D125)
so, when the root is an odd root, you will always get only one number that will produce the radicand.
First, find the scale factor by dividing the first building's real-life height by its model height.
Now, we'll write an equation to find the model height of the second building.
Here is an equation where 
represents the real-life height of a building, 
represents the scale factor, and 
represents the model height of the same building.
Fill in the information we already know.
Divide both sides by .
So, the model height of the second building is 
inches.
