We have JM=ML due to the relation between sides and angles
Set 3x+4 = 5x-16
=> x= 28/5
Substitute back to the equation ( either one would work be cause they are equal) : 3 ( 28/5) +4 = 20.8
We have JL = JM+ML
=> JL = 20.8+20.8 = 41.6
<h2>
Hello!</h2>
The answer is: ![g(x)=-\sqrt[3]{x-1}](https://tex.z-dn.net/?f=g%28x%29%3D-%5Csqrt%5B3%5D%7Bx-1%7D)
<h2>
Why?</h2>
Let's check the roots and the shown point in the graphic (2,-1)
First,
![0=-\sqrt[3]{x-1}\\\\0^{3}=(-\sqrt[3]{x-1})^{3}\\\\0=-(x-1)\\\\x=1](https://tex.z-dn.net/?f=0%3D-%5Csqrt%5B3%5D%7Bx-1%7D%5C%5C%5C%5C0%5E%7B3%7D%3D%28-%5Csqrt%5B3%5D%7Bx-1%7D%29%5E%7B3%7D%5C%5C%5C%5C0%3D-%28x-1%29%5C%5C%5C%5Cx%3D1)
then,
![g(0)=-\sqrt[3]{0-1}\\g(0)=-(-1)\\g(0)=1\\y=1](https://tex.z-dn.net/?f=g%280%29%3D-%5Csqrt%5B3%5D%7B0-1%7D%5C%5Cg%280%29%3D-%28-1%29%5C%5Cg%280%29%3D1%5C%5Cy%3D1)
So, we know that the function intercepts the axis at (1,0) and (0,1), meaning that the function match with the last given option
(
)
Second,
Evaluating the function at (2,-1)
![y=-\sqrt[3]{x-1}\\-1=-\sqrt[3]{2-1}\\-1=-\sqrt[3]{1}\\-1=-(1)\\-1=-1](https://tex.z-dn.net/?f=y%3D-%5Csqrt%5B3%5D%7Bx-1%7D%5C%5C-1%3D-%5Csqrt%5B3%5D%7B2-1%7D%5C%5C-1%3D-%5Csqrt%5B3%5D%7B1%7D%5C%5C-1%3D-%281%29%5C%5C-1%3D-1)
-1=-1
It means that the function passes through the given point.
Hence,
The equation which represents g(x) is ![g(x)=-\sqrt[3]{x-1}](https://tex.z-dn.net/?f=g%28x%29%3D-%5Csqrt%5B3%5D%7Bx-1%7D)
Have a nice day!
Answer:

General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra II</u>
- Distance Formula:

Step-by-step explanation:
<u>Step 1: Define</u>
<em>Find points from graph.</em>
Point A(1, 4)
Point B(-2, -3)
<u>Step 2: Find distance </u><em><u>d</u></em>
Simply plug in the 2 coordinates into the distance formula to find distance<em> d</em>
- Substitute in points [DF]:

- (Parenthesis) Subtract:

- [√Radical] Exponents:

- [√Radical] Add:

- [√Radical] Evaluate:

- Round:

A rectangular school banner
has a length of 44 inches, a perimeter of 156 inches, and an area of 1,496
square inches. the cheerleaders make signs similar to the banner. the length of
a sign is 11 inches. First solve the width of the rectangle:
1496 sq in/ 44 = 34 in
So the sign has also a width
of 34 in and a length of 11 so the area is 34*11 =374 sq in
<span>The perimeter is (34*2)
+(11*2) =90 in</span>