First off, let's convert the decimal to a fraction, notice, we have two decimals, so we'll use in the denominator, a 1 with two zeros then, two decimals, two zeros, thus
now, we know then the ratio dimensions for the new photograph,
![\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\ -------------------------------\\\\ \cfrac{7}{4}\implies \cfrac{4+3}{4}\implies \cfrac{4}{4}+\cfrac{3}{4}\implies 1+\boxed{\cfrac{3}{4}}\impliedby \textit{perimeter is }\frac{3}{4}\textit{ larger} \\\\\\ \stackrel{areas'~ratio}{\cfrac{s^2}{s^2}}\implies \cfrac{3^2}{4^2}\implies \cfrac{9}{16}\impliedby \textit{area is }\frac{9}{16}\textit{ larger than original}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5Ctextit%7Bsimilar%20shape%7D%7D%7B%5Ctextit%7Bsimilar%20shape%7D%7D%5Cqquad%20%5Ccfrac%7Bs%7D%7Bs%7D%3D%5Ccfrac%7B%5Csqrt%7Bs%5E2%7D%7D%7B%5Csqrt%7Bs%5E2%7D%7D%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bs%5E3%7D%7D%7B%5Csqrt%5B3%5D%7Bs%5E3%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Ccfrac%7B7%7D%7B4%7D%5Cimplies%20%5Ccfrac%7B4%2B3%7D%7B4%7D%5Cimplies%20%5Ccfrac%7B4%7D%7B4%7D%2B%5Ccfrac%7B3%7D%7B4%7D%5Cimplies%201%2B%5Cboxed%7B%5Ccfrac%7B3%7D%7B4%7D%7D%5Cimpliedby%20%5Ctextit%7Bperimeter%20is%20%7D%5Cfrac%7B3%7D%7B4%7D%5Ctextit%7B%20larger%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bareas%27~ratio%7D%7B%5Ccfrac%7Bs%5E2%7D%7Bs%5E2%7D%7D%5Cimplies%20%5Ccfrac%7B3%5E2%7D%7B4%5E2%7D%5Cimplies%20%5Ccfrac%7B9%7D%7B16%7D%5Cimpliedby%20%5Ctextit%7Barea%20is%20%7D%5Cfrac%7B9%7D%7B16%7D%5Ctextit%7B%20larger%20than%20original%7D)
I hope this helps have a good day
The x-intercepts for this equation are (-1,0), and (3,0).
Let's break down the problem step by step.
The sum of 4 times a larger integer can be summarized as:
4x
9 times a smaller integer can be summarized as:
9y
We know that we're adding the two expressions because the beginning of the problem has sum. We also know the sum is 3.
So, we get:
4x + 9y = 3
Continuing with the problem, we have the difference between 8 times the larger
8x
We also have 3 times the smaller
3y
This part of the problem mentions the difference between these expressions so we subtract and we know the difference is 27.
8x - 3y = 27
Now, we have both equations so we can solve for either x or y.
4x + 9y = 3
8x - 3y = 27
It's easier to multiply the 4x equation by 2 to have both equations resulting in 8x
(4x + 9y = 3) * 2
We get:
8x +18y = 6
So, we now have:
8x + 18y = 6
8x - 3y = 27
We can subtract the x to get y by itself. We get:
21y = -21
Solve for y and we get:
y = -1
Since we know what y is, we can plug that in and solve for x. It doesn't matter which equation you plug y back in. I'm gonna use the second equation.
8x - 3y = 27
8x - 3 (-1) = 27
8x + 3 = 27
8x = 24
x = 3
So, we now know that
y = -1
x = 3
Hope this helped! If you have any questions then please leave a comment!
Answer:
Graph E: Y= -1/3x +3
Grpah F: Y= 1/2x
Graph G: Y= x-3
Graph H: Y= -x
Step-by-step explanation:
