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lara [203]
3 years ago
8

Ke Shawn is placing a right triangle in the coordinate plane. He knows that the longer leg is twice the length of the shorter le

g. He places the longer leg on the x-axis.
What coordinates should he assign to the third vertex of the triangle?

Mathematics
2 answers:
Artist 52 [7]3 years ago
7 0

the 3rd vertex is on the x axis line which means Y =0

 since the longer leg is twice as long as the shorter leg and the length of the shorter leg is a, then twice the length would be 2a ( 2 times a)


 so the 3rd vertex is (2a,0)

Nutka1998 [239]3 years ago
5 0

Answer:

The third vertex of triangle is (2a,0)

Step-by-step explanation:

We are given the following information in the question:

A right angled triangle such the longer side is twice the length of the shorter side.

We have to find the coordinate of the third vertex of triangle.

Let A(0,a), B(0,0), C(x,0)

Distance between two points with coordinate (x_1,y_1), (x_2,y_2) is given by:

\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}

Now, we are given that the longer side is twice the length of the shorter side.

Thus, we can write:

AC = 2AB\\\sqrt{(x-0)^2 + (0-0)^2}= 2\sqrt{(0-0)^2 + (a-0)^2}\\\pm x = \pm 2a\\\text{Since the point lies on the positive side of x-axis}\\x = 2a

The third vertex of triangle is (2a,0)

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The interquartile range (IQR) for the<br>data:<br>32,7,3,-5,1,8<br>, 10,36 is :​
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Answer:

<em>Your Interquartile range (IQR) would be 19.</em>

Step-by-step explanation:

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3 years ago
Please someone help me with this word problem for my math class T.T
olga55 [171]

Answer:

<u>Perimeter</u>:

= 58 m (approximate)

= 58.2066 or 58.21 m (exact)

<u>Area:</u>

= 208 m² (approximate)

= 210.0006 or 210 m² (exact)

Step-by-step explanation:

Given the following dimensions of a rectangle:

length (L) = \sqrt{252} meters

width (W) = \sqrt{175} meters

The formula for solving the perimeter of a rectangle is:

P  = 2(L + W) or 2L + 2W

The formula for solving the area of a rectangle is:

A = L × W

<h2>Approximate Forms:</h2>

In order to determine the approximate perimeter, we must determine the perfect square that is close to the given dimensions.  

13² = 169

14² = 196

15² = 225

16² = 256

Among the perfect squares provided, 16² = 256 is close to 252 (inside the given radical for the length), and 13² = 169 (inside the given radical for the width).  We can use these values to approximate the perimeter and the area of the rectangle.

P  = 2(L + W)

P = 2(13 + 16)

P = 58 m (approximate)

A = L × W

A = 13 × 16

A = 208 m² (approximate)

<h2>Exact Forms:</h2>

L = \sqrt{252} meters = 15.8745 meters

W = \sqrt{175} meters = 13.2288 meters

P  = 2(L + W)

P = 2(15.8745 + 13.2288)

P = 2(29.1033)

P = 58.2066 or 58.21 m

A = L × W

A = 15.8745 × 13.2288

A = 210.0006 or 210 m²

8 0
3 years ago
Please help me with this I don't really understand it​
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Answer:

Step-by-step explanation:

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