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4 years ago

When the beam makes an angle of 40

1 answer:

3 0

A construction crew wants to hoist a heavy
beam so that it is standing up straight. They
tie a rope to the beam, secure the base, and
pull the rope through a pulley to raise one
end of the beam from the ground. When
the beam makes an angle of 40 degrees with the
ground, the top of the beam is 8 ft above
the ground.
Th e construction site has some telephone
wires crossing it. Th e workers are
concerned that the beam may hit the wires.
When the beam makes an angle of 60 degrees with
the ground, the wires are 2 ft above the top
of the beam. Will the beam clear the wires
on its way to standing up straight?

Math - Steve Thursday, April 16, 2015 at 6:22pmwe see that the length of the beam is

8/sin40 = 12.45 ft

At 60 degrees, the top is

12.45sin60 = 10.78 ft high

So, the wire is 12.78 ft up.

Since the beam is only 12.45 ft long, it will not touch the wires.

You might be interested in

Three runners in a relay race share the 8-mile distance equally. How many miles does each runner travel?

Serggg [28]

Answer: 2 2/3 miles

Step-by-step explanation:

From the question, we are informed that three runners in a relay race share the 8-mile distance equally.

The number of miles that each runner travel will be the total miles divided by number of runners. This will be:

= 8/3

= 2 2/3 miles

4 0

3 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

3 years ago

What's the answer of the this-48.3÷(-7)

goblinko [34]

the answer is 6.9.

When you divide a negative by a negative it is a positive

4 0

4 years ago

Due tomarrow please help!!!!!!!!!!!!!!!!!!!!!!!!! Identify the independent and dependent variables in the following statement.

BlackZzzverrR [31]

The independent variable is the variable that is being changed. In this statement, it would be water flowing flowing in the pool. The dependent variable is what is being changed so that would be the depth of the water.

There, I hope this helps! :)

6 0

3 years ago

A contractor purchases 7 dozen pairs of padded work gloves for $103.32. He incorrectly calculates the unit price at $14.76 per p

kvv77 [185]

Answer:

The correct unit price would be $1.23.

The contractor divided 103.32 by 7, resulting in an error.

Step-by-step explanation:

If we know a dozen equals 12, then 7 dozen looks like:

$7(12)$