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tino4ka555 [31]
3 years ago
5

There are 45 new houses being built in a neighborhood Last month, 1/3 of them were sold. This month, 1/5 of the remaining houses

were sold. How many houses are left be sold?
Mathematics
1 answer:
goldfiish [28.3K]3 years ago
7 0
1/3 of 45 is 15.

So 45-15= 30.

1/5 of 30 is 6.

So 30-6=24.

There are 24 houses left to be sold.
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Sonbull [250]

Answer:

I THINK 96

Step-by-step explanation:

11x11=121

5x5=25

121-25=96

7 0
3 years ago
The area of a rectangle is 1176 square meters. The width of the rectangle is 21 meters. What i the length of the rectangle?
stepan [7]

Answer:

56

Step-by-step explanation:

1176 divided by 21 = 56

3 0
3 years ago
Math help please i will mark brainliest!!!!!
mel-nik [20]

Answer:

The second choice is the correct one.

Step-by-step explanation:

20x^4 - 45x^2

GCF is 5x^2, so the factors are:

5x^2( 4x^2 - 9)

Now 4x^2 - 9 is the differencr of 2 squares:

4x^2 - 9 = (2x + 3)(2x - 3)>

4 0
3 years ago
5x +14+13x-2 i need help on these type of problems please
OleMash [197]

Answer:

18x+12

Step-by-step explanation:

5x+14+13x-2

combine Like terms:

5x+13x

14-2

18x+12

3 0
4 years ago
Read 2 more answers
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.

5 0
4 years ago
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