![\bf ~\hspace{10em}\textit{function transformations} \\\\\\ \begin{array}{llll} f(x)= A( Bx+ C)^2+ D \\\\ f(x)= A\sqrt{ Bx+ C}+ D \\\\ f(x)= A(\mathbb{R})^{ Bx+ C}+ D \end{array}\qquad \qquad \begin{array}{llll} f(x)=\cfrac{1}{A(Bx+C)}+D \\\\\\ f(x)= A sin\left( B x+ C \right)+ D \end{array} \\\\[-0.35em] ~\dotfill\\\\ \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\ \bullet \textit{ flips it upside-down if } A\textit{ is negative}\\ ~~~~~~\textit{reflection over the x-axis}](https://tex.z-dn.net/?f=%5Cbf%20~%5Chspace%7B10em%7D%5Ctextit%7Bfunction%20transformations%7D%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20f%28x%29%3D%20A%28%20Bx%2B%20C%29%5E2%2B%20D%20%5C%5C%5C%5C%20f%28x%29%3D%20A%5Csqrt%7B%20Bx%2B%20C%7D%2B%20D%20%5C%5C%5C%5C%20f%28x%29%3D%20A%28%5Cmathbb%7BR%7D%29%5E%7B%20Bx%2B%20C%7D%2B%20D%20%5Cend%7Barray%7D%5Cqquad%20%5Cqquad%20%5Cbegin%7Barray%7D%7Bllll%7D%20f%28x%29%3D%5Ccfrac%7B1%7D%7BA%28Bx%2BC%29%7D%2BD%20%5C%5C%5C%5C%5C%5C%20f%28x%29%3D%20A%20sin%5Cleft%28%20B%20x%2B%20C%20%5Cright%29%2B%20D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cbullet%20%5Ctextit%7B%20stretches%20or%20shrinks%20horizontally%20by%20%7D%20A%5Ccdot%20B%5C%5C%5C%5C%20%5Cbullet%20%5Ctextit%7B%20flips%20it%20upside-down%20if%20%7D%20A%5Ctextit%7B%20is%20negative%7D%5C%5C%20~~~~~~%5Ctextit%7Breflection%20over%20the%20x-axis%7D)
with that template in mind, let's see
down by 5 units, D = -5
to the left by 4 units, C = +4
Answer:
y<-5
Step-by-step explanation:
y-7<-12
y<-5
Answer:
Approximately 6.4
Step-by-step explanation:
We can use the pythagorean thereom here, that tells us (a^2)+(b^2)=c^2. C is the hypotenuse, the side opposite from the right angle, while a and b are the other sides. We can insert 5 and 4 as a and b, and solve for c
:(5^2)+(4^2)=c^2
:25+16=c^2
:41=c^2
:sqrt(41)=6.4=c (We square rooted both sides. 6.4 is only rounded to the nearest hundredths place.) Hope this helps!
In my opinion I think you would be able to make atleast 2 outfits
