If the length of a rectangle is 3m longer than its width, then:
L=W+3
Is the area really 154^2? Or is it 154m^2? If yes, then:
A=LW
154=(W+3)(W)
154=(W^2+3W)
0=W^2+3W-154
0=(W-11)(W+14)
This means either (W-11) or (W+14) is equal to zero so:
W=11 and W=-14
To find out let's substitute the numbers:
154=(11+3)(11)
154=154
Therefore, the width of the rectangle is 11m
Let that be

Two vertical asymptotes at -1 and 0

If we simply

- Denominator has degree 2
- Numerator should have degree as 2 and coefficient as 3 inorder to get horizontal asymptote y=3 means the quadratic equation should contain 3x²
- But there should be a x intercept at -3 so one zeros should be -3
Find a equation
Find zeros
Horizontal asymptote
So our equation is

Graph attached
Answer:
45.1
Step-by-step explanation:
The original amount is 82
But the wrong amount is 45
So we will have to calculate the percentage for the error
82-45
37
So %error= 37/82 ×100
=3700/82
=45.121~45.1( to nearest tenth)
So the final answer is 45.1%
Q1. Look at the picture.

Q2. Look at the picture.

Q3.
Put the value of x = 2 to the equation 3x + y = 5:

<em>subtract 6 from both sides</em>

Q4.

Substitute (*) to (**):
<em>use distributive property</em>

<em>add 33 to both sides</em>
<em>divide both sides by 11</em>

Put the value of m to (*):


Q5.
w - width
3w - length
24 in - the sum of length and width
The equation:

<em>divide both sides by 4</em>


