Answer:
This idea of reflection correlating with a mirror image is similar in math.
This complete guide to reflecting over the x axis and reflecting over the y axis will provide a step-by-step tutorial on how to perform these translations.
First, let’s start with a reflection geometry definition
Math Definition: Reflection Over the X Axis
A reflection of a point, a line, or a figure in the X axis involved reflecting the image over the x axis to create a mirror image. In this case, the x axis would be called the axis of reflection.
Math Definition: Reflection Over the Y Axis
A reflection of a point, a line, or a figure in the Y axis involved reflecting the image over the Y axis to create a mirror image. In this case, theY axis would be called the axis of reflection.
What is the rule for a reflection across the X axis?
The rule for reflecting over the X axis is to negate the value of the y-coordinate of each point, but leave the x-value the same.
Well, the first line has a slope of -3, and runs through 0, -1.
a line parallel to that one, will have the same exact slope of -3.
now, we know about this other parallel line that it runs through -3,1, and of course, since is parallel, it has a slope of -3
![\bf \begin{array}{ccccccccc} &&x_1&&y_1\\ &&(~ -3 &,& 1~) \end{array} \\\\\\ % slope = m slope = m\implies -3 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-1=-3[x-(-3)] \\\\\\ y-1=-3(x+3) \implies y-1=-3x-9\implies y=-3x-8](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%5C%5C%0A%26%26%28~%20-3%20%26%2C%26%201~%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0A%25%20slope%20%20%3D%20m%0Aslope%20%3D%20%20m%5Cimplies%20-3%0A%5C%5C%5C%5C%5C%5C%0A%25%20point-slope%20intercept%0A%5Cstackrel%7B%5Ctextit%7Bpoint-slope%20form%7D%7D%7By-%20y_1%3D%20m%28x-%20x_1%29%7D%5Cimplies%20y-1%3D-3%5Bx-%28-3%29%5D%0A%5C%5C%5C%5C%5C%5C%0Ay-1%3D-3%28x%2B3%29%0A%5Cimplies%20%0Ay-1%3D-3x-9%5Cimplies%20y%3D-3x-8)
The prime factorization of 10 is 2*5. No exponents required.
This problem can be solved by the chicken rabbits method or you can just do simple algebra.
I.) Chicken and rabbits method
First assume all 110 coins are dimes and none are quarters.
We will have a total value of 11 dollars
Now for each dime we switch out for a quarter, we adds 15 cents to the total value.
18.50-11=7.50 dollars
There are 750/15=50 group of 15 cents in the 7 dollars and 50 cents.
This also meant that we need to switch out 50 dimes for 50 quarters.
So we have 50 quarters.
That first method is very good and very quick once you get the hang of it, now I'm going to show you the algebraic way to solve this.
Let's say there are x dimes and y quarters.
Set up equation
x+y=110
10x+25y=1850
Now solve multiply first equation by 10
10x+10y=1100
subtract
15y=750
y=50
Now we set the numbers of quarters to y so the answer is 50 quarters.
I personally recommend using algebra whenever you can because the practice is very important and you will eventually get really fast at setting up and solving equations. The first method is faster in this case but the second is more generalize, hope it helps.
