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2 years ago

If the length of one leg in this triangle is .find the length of the hypotenuse.

Mathematics

1 answer:

Sonja [21]2 years ago

6 0

By Pythagorean theorem we find that the length of the hypotenuse is $r = \sqrt{2}\cdot l$ .

<h3>How to find the length of the hypotenuse</h3>

In the image attached we have a <em>right</em> triangle with two legs of <em>same</em> length. The hypotenuse of <em>right</em> triangles can be found by Pythagorean theorem, whose form is:

$r = \sqrt{l^{2}+l^{2}}$

$r = \sqrt{2}\cdot l$ , where l is the length of the legs.

By Pythagorean theorem we find that the length of the hypotenuse is $r = \sqrt{2}\cdot l$ .

To learn more on Pythagorean theorem: brainly.com/question/26183488

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The input-output table shows and values that satisfy the equation Y= 4X +21

Hunter-Best [27]

Answer: 41

Explanation: Use the equation. If you put the x value, which is 5, into the equation, you can solve.

4(5)+21

4 x 5 is 20

20+21=41

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3 years ago

What is the Area of this triangle? WILL MARK BRAINLIEST

joja [24]

Answer= 18

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3 years ago

Sally is standing at one corner of a snowy rectangular field that measures 200 ft by 200 ft. She wishes toreach a warm cabin loc

pshichka [43]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The optimal point(i.e the position of P) that will allow Sally to reach the cabin as quickly as possible is $d = 50 \ ft$ from the cabin.

Step-by-step explanation:

From the question we are told that

The speed from the point Sally is standing to the point P is $u = 3 \ ft /s$

The speed from the point P to the cabin is $v = 5 \ ft /s$

Let denote the distance from the bottom left corner to the P be y ft

Generally according to Pythagoras theorem the distance from Sally's first position to point P is mathematically represented as

$s = \sqrt{y^2 + 200^2}$

Generally the distance from that point P to the cabin is mathematically represented as

$d = 200 -y$

Generally the time it takes Sally to cover that distance s is mathematically represented as

$t_1 = \frac{s}{u}$

=> $t_1 = \frac{ \sqrt{y^2 + 200^2}}{3}$

Generally the time it takes Sally to cover that distance d is mathematically represented as

$t_2 = \frac{d}{v}$

=> $t_2 = \frac{ 200 - y }{5}$

So the total time taken is

$t= \frac{ \sqrt{y^2 + 200^2}}{3} + \frac{ 200 - y }{5}$

Generally the minimum time taken with respect to the distance is mathematically evaluated by differentiating and equating the derivative to 0

$\frac{dt}{dy} =\frac{1}{3} \frac{y}{ \sqrt{y^2 + 200^2} } - \frac{1}{5} =0$

=> $\frac{dt}{dy} = \frac{y^2 }{y^2 + 200 ^2} =\frac{9}{25}$

=> $25y^2 = 9y^2 + 9 * 200^2$

=> $16y^2 = 360000$

=> $y = 150$

So the distance from point P to the cabin is

$d = 200 -150$

=> $d = 50 \ ft$

So the optimal point(i.e the position of P) that will allow Sally to reach the cabin as quickly as possible is $d = 50 \ ft$ from the cabin

8 0

3 years ago

The tables below show four ratios of measurements.

Mkey [24]

Answer:D.

Step-by-step explanation:

yards to feet equals 0.3 repeating

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So d is the right answer

3 0

3 years ago

$1240 at 8% compounded semianally for 2 years

larisa86 [58]

Answer:

a=$1446.34

Step-by-step explanation:

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3 years ago

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