Answer:
The sidewalk is 2 feet wide
Step-by-step explanation:
The area of the sidewalk is given as 80 square feet
The length and width of pool
including side walks will be;
10 + 2x and 6 + 2x
To get the area of the side walk
That will be area of total minus area of the pool
That will be ;
(10 + 2x)(6 + 2x) - 10(6) = 80
60 + 12x + 20x + 4x^2 - 60 = 80
4x^2 + 32x -80 = 0
x^2 + 8x -20
x^2 + 10x - 2x - 20 = 0
x(x + 10) -2(x + 10) = 0
(x-2)(x + 10) = 0
x = 2 or -10
since width cannot be negative;
The sidewalk is 2 feet wide
Answer:
1 2/3
Step-by-step explanation:
fraction calc :)
Hello, the answer to your question is B-0.8cm. Hope you have a good day!
<h3>
Answer: B. 2</h3>
How I got that answer:
The data set for Marcus is
2, 3, 5, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 10, 10
The five number summary is
- min = 2
- Q1 = 6
- median = 7
- Q3 = 8
- max = 10
So the interquartile range (IQR) is
IQR = Q3 - Q1
IQR = 8 - 6
IQR = 2
F(x) = (-x)^3
f(-3) = (-(-3)^3 = 3^3 = 27