Ask question

Ask question

All categories

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?

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.

You might be interested in

2.4 liters of water is poured into a pitcher that now contains 10.1 liters of water. Which equation represents this?

Alenkinab [10]

I couldn’t figure out d answer to This hope this helped

8 0

2 years ago

What is the value of y in the product of powers below?

Marat540 [252]

Answer:

y =0

Step-by-step explanation:

From the equation;

8³ × 8⁻⁵×8^y = 8⁻²= 1/8²

From the laws of indices;

aⁿ×aⁿ = a^2n

Therefore;

8³ × 8⁻⁵×8^y = 8^(3+-5+y)

8^(-2+y) = 8^-2 ; but the bases are the same and thus the exponents are the same;

-2 + y = -2

y = 0

8 0

3 years ago

Read 2 more answers

What is one factor of 20x2 + 6x − 8

Furkat [3]

You can factor out a 2 from all. Idk if that helps but here ya go. 10x+3x-4

8 0

3 years ago

Read 2 more answers

In a sale a clothes shop reduces its price by 40%.a shirt usually costs £20 how much is it in sale

irakobra [83]

If it is 40% off, then u are paying 60%

0.60(20) = 12 <== on sale, it is 12 pounds

5 0

3 years ago

Consider w=sqrrt2/2(cos(225°) + isin(225°)) and z = 1(cos(60°) + isin(60°)). What is w+ z expressed in rectangular form?

SVETLANKA909090 [29]

Answer:

Option (3)

Step-by-step explanation:

w = $\frac{\sqrt{2}}{2}[\text{cos}(225) + i\text{sin}(225)]$

Since, cos(225) = cos(180 + 45)

= -cos(45) [Since, cos(180 + θ) = -cosθ]

= - $\frac{\sqrt{2}}{2}$

sin(225) = sin(180 + 45)

= -sin(45)

= - $\frac{\sqrt{2}}{2}$

Therefore, w = $\frac{\sqrt{2}}{2}[-\frac{\sqrt{2}}{2}+i(-\frac{\sqrt{2}}{2})]$

= $-\frac{2}{4}(1+i)$

= $-\frac{1}{2}(1+i)$

z = 1[cos(60) + i(sin(60)]

= $[\frac{1}{2}+i(\frac{\sqrt{3}}{2})$

= $\frac{1}{2}(1+i\sqrt{3})$

Now (w + z) = $-\frac{1}{2}(1+i)+\frac{1}{2}(1+i\sqrt{3})$

= $-\frac{1}{2}-\frac{i}{2}+\frac{1}{2}+i\frac{\sqrt{3}}{2}$

= $\frac{(i\sqrt{3}-i)}{2}$

= $\frac{(\sqrt{3}-1)i}{2}$

Therefore, Option (3) will be the correct option.

3 0

3 years ago

Read 2 more answers