We can see that there are 5 CDs, each of radius 9 cm
<u>Area occupied by 1 disc:</u>
Area of a circle = πr²
Area of disc = π(9)²
Area of disc = 3.14 * 81 = 254 cm²
<u>Area occupied by 5 discs:</u>
Area occupied by 5 discs = Area occupied by 1 disc * 5
Area occupied by 5 discs = 254 * 5
Area occupied by 5 discs = 1270 cm²
Step-by-step explanation:
11.
Area of rectangle = 408 in²
Length = 3x-2 and width = x
The area of a rectangle is given by :
A = lb
(3x-2)x = 408
x = 12 inch and x = -11.33 in
Neglecting negative value,
The width of rectangle = 12 inch
The length of rectangle = 3x-2 = 3(12)-2 = 34 inch
12.
Base of triangle = 2x+2
Height = x
The area of a triangle is given by :
x = 24 cm and x = -25 cm
Neglecting negative value, x = 24 cm
Base = 2x+2 = 2(24)+2 = 50 cm
Hence, this is the required solution.
Answer:
Your answer is 3 quarters, 2 nickels, and 4 pennies
Step-by-step explanation:
2 quarters plus 6 nickels plus 9 pennies is equal to 89 cents. So if we subtract 75(3 quarters) we get 14 which is less than 25 therefore we must only use nickels and pennies from now on. If we take 14 and subtract 10(2 nickels) we get 4 which less than 5 leaving us only to use pennies, which one a piece so it must be 4 pennies. Hope this helps!
Answer:
5X + 8Y >= 300; intersection at (-20, 50)
Step-by-step explanation:
let t = work hours
0 < t < 30
X = time lawn mowing
Y = time babysitting.
X + Y < 30
5X + 8Y >= 300
We could solve...
X < 30 - Y
5(30 - Y) + 8Y >=300
150 - 5Y + 8Y >= 300
3Y >=150
Y >=50
then X < -20
intersection at (-20, 50)
C) average of a data set
You add all of the data and divide by the number of data
