Alright so, let’s break this down.
To find the first coordinate you need to solve the first equation
To do that you can do
6+9=15
So that mean 3x has to be equal to 15
You divide 15 by 3
The answer to that is the solution to the first equation, or the first coordinate.
For the second one, you know x has to be negative because the solution is negative
2x is going to have to equal -20
-20/2 gets you the solution
Remember that it will be negative
You now have both of the coordinates
You take coordinate A and technically add coordinate B to it
You do that because B is negative and A is positive
Then you have the answer to c
I am pretty sure you mistyped the rule.
The rule (x + 7, y - 3) moves the original coordinate right 7 units and down 3 units
The area of a rhombus is 1/2 times the lengths of the diagonals
A = (1/2)(AC)(BD)
Since the diagonals bisect each other, and ed = 8, BD must be 16. Substituting in this for BD and the given area (168) we get the following:
168 = (1/2)(AC)16
168 = 8(AC)
21 = AC
Answer:
![\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B1.%5C%20f%5E%7B-1%7D%28x%29%3D4%5Clog%28x%5Csqrt%5B4%5D2%29%7D%5C%5C%5C%5C%5Cboxed%7B2.%5C%20f%5E%7B-1%7D%28x%29%3D%5Clog%28x%5E5%2B5%29%7D%5C%5C%5C%5C%5Cboxed%7B3.%5C%20f%5E%7B-1%7D%28x%29%3D%5Csqrt%7B4%5E%7Bx-1%7D%7D%7D)
Step-by-step explanation:
![\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)](https://tex.z-dn.net/?f=%5Clog_55%5E%7B%5Cfrac%7B1%7D%7B4%7Dy%7D%3D%5Clog_5%5Cleft%282%5E%5Cfrac%7B1%7D%7B4%7Dx%5Cright%29%5Cqquad%5Ctext%7Buse%7D%5C%20a%5E%5Cfrac%7B1%7D%7Bn%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5C%5C%5C%5C%5Cdfrac%7B1%7D%7B4%7Dy%3D%5Clog%28x%5Csqrt%5B4%5D2%29%5Cqquad%5Ctext%7Bmultiply%20both%20sides%20by%204%7D%5C%5C%5C%5Cy%3D4%5Clog%28x%5Csqrt%5B4%5D2%29)
![--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)](https://tex.z-dn.net/?f=--------------------------%5C%5C2.%5C%5Cy%3D%2810%5Ex-5%29%5E%5Cfrac%7B1%7D%7B5%7D%5C%5C%5C%5C%5Ctext%7BExchange%20x%20and%20y.%20Solve%20for%20y%3A%7D%5C%5C%5C%5C%2810%5Ey-5%29%5E%5Cfrac%7B1%7D%7B5%7D%3Dx%5Cqquad%5Ctext%7B5%20power%20of%20both%20sides%7D%5C%5C%5C%5C%5Cbigg%5B%2810%5Ey-5%29%5E%5Cfrac%7B1%7D%7B5%7D%5Cbigg%5D%5E5%3Dx%5E5%5Cqquad%5Ctext%7Buse%7D%5C%20%28a%5En%29%5Em%3Da%5E%7Bnm%7D%5C%5C%5C%5C%2810%5Ey-5%29%5E%7B%5Cfrac%7B1%7D%7B5%7D%5Ccdot5%7D%3Dx%5E5%5C%5C%5C%5C10%5Ey-5%3Dx%5E5%5Cqquad%5Ctext%7Badd%205%20to%20both%20sides%7D%5C%5C%5C%5C10%5Ey%3Dx%5E5%2B5%5Cqquad%5Clog%5C%20%5Ctext%7Bof%20both%20sides%7D%5C%5C%5C%5C%5Clog10%5Ey%3D%5Clog%28x%5E5%2B5%29%5CRightarrow%20y%3D%5Clog%28x%5E5%2B5%29)
Answer:
13 - 2 = 11
Step-by-step explanation:
