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Andru [333]
3 years ago
15

Show the work please.

Mathematics
1 answer:
sladkih [1.3K]3 years ago
3 0

In any given right triangle, the Pythagorean Theorem can be used to show that it is a right triangle.

The Pythagorean Theorem is a^2+b^2=c^2. In a right triangle, a and b would be the shorter legs of the triangle, while c would be the hypotenuse.

So for this problem, you would plug in the numbers in the order that they are listed to see if it is a right triangle.

F would be: 2^2+4^2=7^2. In this case, the sides are not equal.

G would be: 6^2+8^2=10^2. In this case, 100=100. So this is a right triangle.

H would be: 4^2+9^2=12^2. The sides are not equal.

J would be: 5^2+10^2=15^2. The sides are not equal.

Your answer would be G, since the sides are equal. Hope this helps! :)

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What is the answer to the question down below
ratelena [41]

We conclude that the measure of that angle is 150 degrees.

<h3>How to get the measure of angle BFE?</h3>

Notice that  the measure of angle ∠BFE will be equal to:

∠BFE = ∠BFC + ∠CFD + ∠DFE

Where:

∠CFD = 90°

∠DFE = 30°

And we will have that:

∠AFC = 180° - 90° - 30° = 60°

And we know that BF bisects that angle, then:

∠BFC = 30°

So we conclude that:

∠BFE = ∠BFC + ∠CFD + ∠DFE = 30° + 90° + 30° = 150°

We conclude that the measure of that angle is 150 degrees.

If you want to learn more about angles:

brainly.com/question/17972372

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5 0
2 years ago
What are the first five terms
tangare [24]

Answer:

Its the first choice.

Step-by-step explanation:

an = the nth term. so:

a1 = 6(1) + 1 = 7

a2 = 6(2) + 1 = 13

a3 = 6(3) + 1 = 19

The common difference is 6.

5 0
3 years ago
Read 2 more answers
I need help with this problem
erik [133]
I think the answer is 400 square centimetres
8 0
3 years ago
Two circles, one of radius 5 inches, the other of radius 2 inches, are tangent at point P. Two bugs start crawling at the same t
lawyer [7]

Answer:

  40 minutes

Step-by-step explanation:

The circumference of the larger circle is ...

  C = 2πr = 2π(5 in) = 10π in

The bug navigates the circumference at 3π in/min, so will take

  time = distance/speed = (10π in)/(3π in/min) = 10/3 min

to travel once around.

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The circumference of the smaller circle is ...

  C = 2πr = 2π(2 in) = 4π in

The bug navigates this circumference at 2.5π in/min, so will take

  (4π in)/(2.5π in/min) = 8/5 min

to travel once around.

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The bugs will meet at a time that is the least common multiple of these times. Both can be expressed in 15ths of a minute as ...

  {50/15, 24/15}

Then the LCM of these will be ...

  (1/15)LCM(50, 24) = (1/15)(50×24)/GCD(50, 24) = 1200/30 = 40

It will be 40 minutes before the bugs next meet at point P.

___

A graphing calculator can make use of the mod function to show when the bugs meet at point P (total is displacement of the two bugs from P is zero). It shows the meeting occurs after 40 minutes.

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150-35 is 115 dollers!!

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