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rosijanka [135]
3 years ago
13

a hotel has 260 rooms- some single and some doubles. The singles cost $69 per night and the doubles cost $85 per night. Because

of a math teacher's convention, all the hotel rooms are occupied. The sales for this night are $21,076. How many singles are there
Mathematics
1 answer:
luda_lava [24]3 years ago
6 0
x-single\ rooms\\260-x-double\ rooms\\\\69x+85(260-x)=21076\\69x+22100-85x=21076\\-16x=21076-22100\\-16x=-1024\ \ \ \ \ |:-16\\x=64\\\\There\ are\ 64\ single\ rooms.
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SECOND TIMES THE CHARM.. please help
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Answer:

y = 5x.

Step-by-step explanation:

Every y value is the corresponding x value times 5. (For example, 3 x 5 = 15, 5 x 5 =25)

<em>Hope this helps!</em>

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What is the volume of a solid if the height is 4in and the base is 18 inches squared
zubka84 [21]
4in x 18in^2= 72 inches cubed
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Two competitive neighbours build rectangular pools that cover the same area but are different shapes. Pool A has a width of (x +
GenaCL600 [577]

<u>Answer: </u>

a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

b) Area of pool A is equal to area of pool B equal to 24.44 meters.

<u> Solution: </u>

Let’s first calculate area of pool A .

Given that width of the pool A = (x+3)  

Length of the pool A is 3 meter longer than its width.

So length of pool A = (x+3) + 3 =(x + 6)

Area of rectangle = length x width

So area of pool A =(x+6) (x+3)        ------(1)

Let’s calculate area of pool B

Given that length of pool B is double of width of pool A.

So length of pool B = 2(x+3) =(2x + 6) m

Width of pool B is 4 meter shorter than its length,

So width of pool B = (2x +6 ) – 4 = 2x + 2

Area of rectangle = length x width

So area of pool B =(2x+6)(2x+2)        ------(2)

Since area of pool A is equal to area of pool B, so from equation (1) and (2)

(x+6) (x+3) =(2x+6) (2x+2)    

On solving above equation for x    

(x+6) (x+3) =2(x+3) (2x+2)  

x+6 = 4x + 4    

x-4x = 4 – 6

x = \frac{2}{3}

Dimension of pool A

Length = x+6 = (\frac{2}{3}) +6 = 6.667m

Width = x +3 = (\frac{2}{3}) +3 = 3.667m

Dimension of pool B

Length = 2x +6 = 2(\frac{2}{3}) + 6 = \frac{22}{3} = 7.333m

Width = 2x + 2 = 2(\frac{2}{3}) + 2 = \frac{10}{3} = 3.333m

Verifying the area:

Area of pool A = (\frac{20}{3}) x (\frac{11}{3}) = \frac{220}{9} = 24.44 meter

Area of pool B = (\frac{22}{3}) x (\frac{10}{3}) = \frac{220}{9} = 24.44 meter

Summarizing the results:

(a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

(b)Area of pool A is equal to Area of pool B equal to 24.44 meters.

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Ive been trying to figure out this problem, I’m not sure if each fruit is measures per pound. Please help!
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Answer:

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Step-by-step explanation:

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How many fifths are the in 60/100
tankabanditka [31]
In order to calculate that, we need to divide it by 1/5 as follows:
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So, there are "3" 1/5ths in 60/100

Hope this helps!
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