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fgiga [73]
3 years ago
12

A Stability cable is attached to the top of an electricity pole and then stretched to the ground.The pole is 4.5m tall while it

he cable has a length of 6m.How far from the base of the pole is the cable attached to the ground

Mathematics
1 answer:
valkas [14]3 years ago
7 0

Answer: 3.96 meters

Step-by-step explanation:

Draw a Right triangle as the one shown in the image attached, where "x" is the distance from the base of the pole to the point in which the cable is attached to the ground.

In this case you need to use the Pythagorean Theorem. According to this theorem:

a^2=b^2+c^2

Where "a" is the hypotenuse and "b" and "c" are the legs of the Right triangle.

If you solve for one of the legs, you get:

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

In this case, you can identify that:

a=6\ m\\\\b=4.5\ m\\\\c=x

Substituting values, you get the following result:

c=\sqrt{(6\ m)^2-(4.5\ m)^2}\\\\c=3.96\ m

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Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.
frutty [35]
<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

2^n\leq 2^{n+1}-2^{n-1}-1

where n is a positive integer.

  • Let us take n=1

then we have:

2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2

Hence, the result is true for n=1.

  • Let us assume that the result is true for n=k

i.e.

2^k\leq 2^{k+1}-2^{k-1}-1

  • Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u>  2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Let us take n=k+1

Hence, we have:

2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)

( Since, the result was true for n=k )

Hence, we have:

2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2

Also, we know that:

-2

(

Since, for n=k+1 being a positive integer we have:

2^{(k+1)+1}-2^{(k+1)-1}>0  )

Hence, we have finally,

2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

2^n\leq 2^{n+1}-2^{n-1}-1  where n is a positive integer.

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Answer:

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Step-by-step explanation:

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