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3 years ago

7

Mr. Krogman usually prices his umbrellas for $8 each. However, on rainy days he increases them by 75%. How much does he charge f

or an umbrella on a rainy day?

1 answer:

Serhud [2]3 years ago

7 0

Answer:

The answer is 12.5.

Step-by-step explanation:

Step 1 : 8 dollars times 75% is 4.5

Step 2: Add 4.5 plus 8 dollars

Step 3: 12.5

On a rainy day an umbrella costs 12.50 dollars in total.

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Chase collects 27.34 for a fundraiser.Sydney collects 9.83 more than chase.Ally collects 4 times as much as sydney.How much mone

Vinil7 [7]

Collection of Chase = 27.34

Collection of Sydney = 9.83+27.34 = 37.17

Let ally's collection be = x

As Ally collects 4 times as much as Sydney so,

x= 4*(37.17) =148.68

Hence, Ally collects 148.68 units

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3 years ago

120/210 *60 (give the answer in hour and minutes)

Katyanochek1 [597]

120/210 = 0.57
0.57 x 60 = 34.2

34 hours and 12 minutes

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3 years ago

A rectangular carpet is x+5 feet wide and 3x-2 feet long what is the area in square feet of the carpet in terms of x

siniylev [52]

Formula for area:
Area = Wide x Length

Wide: x + 5
Long: 3x - 2

Area = (x + 5) x (3x - 2)
Area = 3x² - 2x + 15x - 10
Area = (3x² + 13x - 10) square feet

7 0

3 years ago

The region bounded by y=x^2+1, y=x, x=-1, x=2 with square cross sections perpendicular to the x-axis.

VLD [36.1K]

Answer:

The bounded area is 5 + 5/6 square units. (or 35/6 square units)

Step-by-step explanation:

Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)

Such that f(x) > g(x) in the given interval.

This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).

We want to find the area bounded by:

f(x) = y = x^2 + 1

g(x) = y = x

x = -1

x = 2

To find this area, we need to f(x) - g(x) between x = -1 and x = 2

This is:

$\int\limits^2_{-1} {(f(x) - g(x))} \, dx$

$\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx$

We know that:

$\int\limits^{}_{} {x} \, dx = \frac{x^2}{2}$

$\int\limits^{}_{} {1} \, dx = x$

$\int\limits^{}_{} {x^2} \, dx = \frac{x^3}{3}$

Then our integral is:

$\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx = (\frac{2^3}{2} + 2 - \frac{2^2}{2}) - (\frac{(-1)^3}{3} + (-1) - \frac{(-1)^2}{2} )$

The right side is equal to:

$(4 + 2 - 2) - ( -1/3 - 1 - 1/2) = 4 + 1/3 + 1 + 1/2 = 5 + 2/6 + 3/6 = 5 + 5/6$

The bounded area is 5 + 5/6 square units.

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2 years ago

After being dropped a certain ball always bounces back to 2/5 of the height of its previous bounce. After the first bounce it re

mylen [45]

After the second, it will be 50, after the third it will be 20, and after the fourth it will be 8. The answer is C
Hope this helps!

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3 years ago