<h3><u>Question:</u></h3>
The perimeter of a rectangle is 34 units. Its width W is 6.5 units.
Write an equation to represent the perimeter in terms of the length L, and find the value of L
<h3><u>Answer:</u></h3>
The length of rectangle is 10.5 units
<h3><u>
Solution:</u></h3>
Given that,
Perimeter of rectangle = 34 units
Width of rectangle = 6.5 units
Let "L" be the length of rectangle
<em><u>The perimeter of rectangle is given by formula:</u></em>
Perimeter = 2(length + width)
<em><u>Substituting the values we get,</u></em>
![34 = 2(L + 6.5)](https://tex.z-dn.net/?f=34%20%3D%202%28L%20%2B%206.5%29)
Thus the equation is found
<em><u>Solve for "L"</u></em>
![L + 6.5 = \frac{34}{2}\\\\L + 6.5 = 17\\\\L = 17 - 6.5\\\\L =10.5](https://tex.z-dn.net/?f=L%20%2B%206.5%20%3D%20%5Cfrac%7B34%7D%7B2%7D%5C%5C%5C%5CL%20%2B%206.5%20%3D%2017%5C%5C%5C%5CL%20%3D%2017%20-%206.5%5C%5C%5C%5CL%20%3D10.5)
Thus length of rectangle is 10.5 units
Answer:
The answer would be 20 weeks
Step-by-step explanation:
Because, if she makes $5 per week you would just divide 100 by 5 and get 20 and to check you would multiply your answer to see if it equals the amount he/she would want to amount to. Hope this helps! Happy learning!
Answer:
The answer is below
Step-by-step explanation:
The formula m = (12,000 + 12,000rt)/12t gives Keri's monthly loan payment, where r is the annual interest rate and t is the length of the loan, in years. Keri decides that she can afford, at most, a $275 monthly car payment. Give an example of an interest rate greater than 0% and a loan length that would result in a car payment Keri could afford. Provide support for your answer.
Answer: Let us assume an annual interest rate (r) = 10% = 0.1. The maximum monthly payment (m) Keri can afford is $275. i.e. m ≤ $275. Using the monthly loan payment formula, we can calculate a loan length that would result in a car payment Keri could afford.
![m=\frac{12000+12000rt}{12t}\\ but\ m\leq275, \ and \ r=10\%=0.1\\275= \frac{12000+12000(0.1)t}{12t}\\275= \frac{12000}{12t} +\frac{12000(0.1)}{12t}\\275= \frac{1000}{t} + 100\\275-100= \frac{1000}{t} \\175= \frac{1000}{t} \\175t = 1000\\t= \frac{1000}{175}\\ t=5.72\ years](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B12000%2B12000rt%7D%7B12t%7D%5C%5C%20but%5C%20m%5Cleq275%2C%20%5C%20and%20%5C%20r%3D10%5C%25%3D0.1%5C%5C275%3D%20%5Cfrac%7B12000%2B12000%280.1%29t%7D%7B12t%7D%5C%5C275%3D%20%5Cfrac%7B12000%7D%7B12t%7D%20%2B%5Cfrac%7B12000%280.1%29%7D%7B12t%7D%5C%5C275%3D%20%5Cfrac%7B1000%7D%7Bt%7D%20%2B%20100%5C%5C275-100%3D%20%5Cfrac%7B1000%7D%7Bt%7D%20%5C%5C175%3D%20%5Cfrac%7B1000%7D%7Bt%7D%20%5C%5C175t%20%3D%201000%5C%5Ct%3D%20%5Cfrac%7B1000%7D%7B175%7D%5C%5C%20t%3D5.72%5C%20years)
The loan must be at least for 5.72 years for an annual interest rate (r) of 10%