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sweet [91]
3 years ago
15

How long is a wire reaching from the top of a 12-foot pole to a point 6 feet from the base of the pole?

Mathematics
2 answers:
miskamm [114]3 years ago
8 0
<span> Using Pythagoras' theorem it is 6 times the sq rt of 5 which is about 13.416 </span><span>feet to 3 decimal places  </span>
Alex17521 [72]3 years ago
4 0
The pole is 12 foot long.

The wire from the top to the ground, 6feet from the pole, forms a right angle triangle.

By Pythagoras' Theorem:

x² = 12² + 6²

x² = 144 + 36

x² = 180          Take the square root of both sides.

x  = √180              Use a calculator

x ≈ 13.42 feet.

Length of wire ≈ 13.42 feet.
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3 years ago
The measure of two supplementary angles are in the ratio 4:11 what is the measure of the larger angle
vichka [17]

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8 0
2 years ago
What is the quotient? Negative StartFraction 3 over 8 EndFraction divided by negative one-fourth
anyanavicka [17]

Answer:

\dfrac{3}{2}

Step-by-step explanation:

We are given 2 fractions.

1st fraction is Negative StartFraction 3 over 8

i.e.

-\dfrac{3}{8}

2nd fraction is negative one-fourth

i.e.

-\dfrac{1}{4}

We have to find the quotient when 1st fraction is divided by 2nd fraction.

<u>Definition</u> of quotient is given as:

Quotient is the result obtained when one number is divided by other number.

-\dfrac{3}{8} \div -\dfrac{1}{4} \text{ is to be calculated.}

\div is converted to \times and the 2nd fraction is reversed i.e.

if the fraction is \frac{p}{q} it becomes \frac{q}{p} and the sign \div is changed to \times.

Solving above:

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6 0
3 years ago
Pratap puri rowed 10 miles down a river in 2 ​hours, but the return trip took him 2 and one half hours. Find the rate pratap can
N76 [4]

Answer:

<em>Rate of Pratap in still water is 4.5 miles/hour and rate of current is 0.5 miles/hour.</em>

Step-by-step explanation:

Pratap Puri rowed 10 miles down a river in 2 ​hours, but the return trip took him 2.5 hours.

We know that, Speed = \frac{Distance}{Time}

So, the <u>speed of Pratap with the current</u> will be:  (\frac{10}{2})miles/hour = 5 miles/hour

and the <u>speed of Pratap against the current</u> will be:  (\frac{10}{2.5})miles/hour = 4 miles/hour.

Suppose, the rate of Pratap in still water is x and the rate of current is y.

So, the equations will be........

x+y= 5 .............................. (1)\\ \\ x-y=4 .............................. (2)

Adding equation (1) and (2) , we will get......

2x=9\\ \\ x=\frac{9}{2}= 4.5

Now, plugging this x=4.5 into equation (1), we will get.....

4.5+y=5\\ \\ y=5-4.5 =0.5

Thus, Pratap can row at 4.5 miles per hour in still water and the rate of the current is 0.5 miles/hour.

6 0
3 years ago
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