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pashok25 [27]
3 years ago
6

Matt can walk 500 m from his home on King Street, and then 900m along Queen Street to go to his school. He can also walk across

a diagonally across a park from his home to his school. Assuming King Street and Queen Street meet at right angles, how far does he walk across the park to go to school?
Mathematics
1 answer:
brilliants [131]3 years ago
7 0

Answer:

He walks 1030 meters if he walks across the park.

Step-by-step explanation:

In order to solve this question we will need to know that c^{2} = a^{2} + b^{2} (were c is the length of the hypotenuse (the diagonal line) of a right angle triangle, and a and b are the legs (the sides that form a right angle). So them means that.....

Let "c" be the distance he will need to walk if he was going to go through the park

Let "a" be the distance he walks on King Street

Let "b " be the distance he walks on Queen Street, then.....

c^{2} = a^{2} + b^{2}

c = \sqrt{a^{2} + b^{2} }

(Now plug in the values of a and b and get.....)

c = \sqrt{500^{2} + 900^{2}  }

c = \sqrt{1060000 }

c = 1029.563014............

So I would assume that you have to round to the nearest meter. And as a result we get 1030 meter.

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Vinvika [58]

Answer:

Option 2

Step-by-step explanation:

Coordinates of point on a graph is given by (x, y).

Where x = distance of the point from origin on x-axis.

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Therefore, coordinates of points given on graph will be,

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Option (2) will be the correct option.

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Answer:

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7 0
3 years ago
Todd is 3 years older than his brother Jack. If Jack is x years old and Todd is y years
o-na [289]
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4 0
4 years ago
Read 2 more answers
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uysha [10]

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Evgen [1.6K]

Answer:

4(x+3.5)^{2}=0

Step-by-step explanation:

we have

4x^{2}+28x+49=0

step 1

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(4x^{2}+28x)=-49

step 2

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step 3

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3 years ago
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