Asked by cool
1. The area of a rectangular garden in square feet is x^2-9x-400. What is the width and length of the garden?
*I factored this one and got (x+16)(x-25), but I don't know which one is the width and which one is the length.
2. A rectangular pool is 15 ft wide and 40 ft long. The pool is surrounded by a walkway. the walkway is the same width all the way around the pool. The total area of the walkway is 456 square feet. How wide is the walkway?
*I'm not sure how to set up the equation, but is it 456=(15+2x)(40+2x)-(15)(40)?
*I factored this one and got (x+16)(x-25), but I don't know which one is the width and which one is the length.
2. A rectangular pool is 15 ft wide and 40 ft long. The pool is surrounded by a walkway. the walkway is the same width all the way around the pool. The total area of the walkway is 456 square feet. How wide is the walkway?
*I'm not sure how to set up the equation, but is it 456=(15+2x)(40+2x)-(15)(40)?
Answers
Answered by
cool
*I'm not sure how to solve these. Can you please help me?
The problem asks you to find the value of k that wuold make x^2-kx+64 a perfect square trinomial. You said k=12. Your friend said k=6. Who is correct? What mistake was made by the incorrect person?
You work at a restaurant whose weekly profit is given by the formula p=-c^2+14c+800, where c is the average price of the food, in dollars. The manager wants to add delivery service, which will cost the restaruant d=5c+300 per week.
Find the highest average price d the restaurant can sell its food at and still make a profit if they add delivery.
What will the weekly profit p be if the restaurant sells its food at this average price and doesn't offer delivery
The problem asks you to find the value of k that wuold make x^2-kx+64 a perfect square trinomial. You said k=12. Your friend said k=6. Who is correct? What mistake was made by the incorrect person?
You work at a restaurant whose weekly profit is given by the formula p=-c^2+14c+800, where c is the average price of the food, in dollars. The manager wants to add delivery service, which will cost the restaruant d=5c+300 per week.
Find the highest average price d the restaurant can sell its food at and still make a profit if they add delivery.
What will the weekly profit p be if the restaurant sells its food at this average price and doesn't offer delivery
Answered by
Reiny
#1, Usually, we consider the length to be the larger number.
so comparing x+16 and x-25, we have 3 cases:
a) they are equal: x+16 = x-25 --> 16 = -25, which is false
b) x+16 < x-25 --> 16 < -25 , which is also false
c) x+16 > x-25, ---> 16 > -25, which is true, so
x+16 is the length, and x-25 is the width
#2, your equation is correct, all you have to do is solve for x
#3 , (x-a)^2 = x^2 - 2ax + a^2
so for x^2-kx+64 , a^2 = 64, then a = 8 and
-kx = -2ax
kx = 2(8)x
k = 16
#4
profit = -c^2+14c+800 + 5c+300
= -c^2 + 19c + 1100
= -(c^2 - 19c - 1100)
= -(c - 44)(c + 25)
break-even is when -(c - 44)(c + 25) = 0
or (c - 44)(c + 25) = 0
c = 44 or x = -25 , but c is price of food, thus c > 0
Does that help in reaching a conclusion?
Try a value above 44 and a value of c below 44 and find the profit using the factored equation for profit.
I am sure you can now find how to do the last problem.
so comparing x+16 and x-25, we have 3 cases:
a) they are equal: x+16 = x-25 --> 16 = -25, which is false
b) x+16 < x-25 --> 16 < -25 , which is also false
c) x+16 > x-25, ---> 16 > -25, which is true, so
x+16 is the length, and x-25 is the width
#2, your equation is correct, all you have to do is solve for x
#3 , (x-a)^2 = x^2 - 2ax + a^2
so for x^2-kx+64 , a^2 = 64, then a = 8 and
-kx = -2ax
kx = 2(8)x
k = 16
#4
profit = -c^2+14c+800 + 5c+300
= -c^2 + 19c + 1100
= -(c^2 - 19c - 1100)
= -(c - 44)(c + 25)
break-even is when -(c - 44)(c + 25) = 0
or (c - 44)(c + 25) = 0
c = 44 or x = -25 , but c is price of food, thus c > 0
Does that help in reaching a conclusion?
Try a value above 44 and a value of c below 44 and find the profit using the factored equation for profit.
I am sure you can now find how to do the last problem.
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