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maw [93]
2 years ago
13

On a treasure map, a X is placed at point (9, -7). if the pirates are currently on point (9,3), how many units away are they?​

Mathematics
2 answers:
Alex_Xolod [135]2 years ago
8 0

Answer:

They are 10 units away from the treasure

Step-by-step explanation:

Notice that the points (9, -7) and (9, 3) are both located at the same x-value (9) while their y-values are in one case 3 units above the x-axis, and in the other 7 units BELOW the x-axis. Therefore thy differ by 3 - (-7) units , which totals 10 units.

kozerog [31]2 years ago
4 0
The answer to this question is 10 units away
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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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riadik2000 [5.3K]

Answer:

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