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3 years ago

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A construction crane lifts a bucket of sand originally weighing 145 lbs at a constant rate. Sand is lost from the bucket at a co

nstant rate of .5lbs/ft. How much work is done in lifting the sand 80ft?

1 answer:

insens350 [35]3 years ago

4 0

Answer: $10,000\ lb.ft$

Step-by-step explanation:

Given

Initial weight of the bucket is $145\ lb$

It is lifted at constant rate and rate of sand escaping is $0.5\ lb/ft$

At any height weight of the sand is $w(h)=145-0.5h$

Work done is given by the product of applied force and displacement or the area under weight-displacement graph

from the figure area is given by

$\Rightarrow W=\int_{0}^{80}\left ( 145-0.5h \right )dh\\\\\Rightarrow W=\left | 145h-\dfrac{0.5h^2}{2} \right |_0^{80}\\\\\Rightarrow W=\left [ 145\times 80-\dfrac{0.5(80))^2}{2} \right ]-0\\\\\Rightarrow W=11,600-1600\\\\\Rightarrow W=10,000\ lb.ft$

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Please help!!!!!!!!!!

pogonyaev

Answer:

By hypotenuse - side test (HL) the two triangles are congruent.

Step-by-step explanation:

In ∆ABC and ∆DCB

i) angleABC = angleDCB.....(each 90°)

ii) BC = BC .....( common side)

iii) AC = DB.....(given)

therefore by hypo-side test ∆ABC congruent ∆DCB

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3 years ago

Read 2 more answers

H=4(x+3y)+2 make x the subject

blagie [28]

H = 4(x + 3y) + 2
H = 4x + 12y + 2
H - 12y - 2 = 4x
(H - 12y - 2) / 4 = x or 1/4H - 3y - 1/2 = x

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3 years ago

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Last year, Kevin had 20,000 to invest. He invested some of it in an account that paid 6% simple interest per year, and he invest

tester [92]

$13,000 was invested at 6% whereas $7,000 was invested at 10% using simple interest approach

What is simple interest?

Simple interest is determined as the amount invested multiplied by the interest rate and the number of years that investment lasts.

I=PRT

I=interest

P=principal

R=rate of return

T=time

Let X be the amount invested at 6%

I at 6%=6%*X*1

I=0.06X

The amount invested at 10% is 20,000-X

total interest=0.06X+0.10*(20000-X)

total interest=1,480

1480=0.06X+0.10*(20000-X)

1480=0.06X+2000-0.10X

1480-2000=0.06X-0.10X

-520=-0.04X

X=-520/-0.04

X=13,000

13,000 was invested at 6%

amount invested at 10%=20,000-X

amount invested at 10%=20,000-13,000

amount invested at 10%=7,000

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1 year ago

Arrange these functions from the greatest to the least value based on the average rate of change in the specified interval.

Romashka [77]

By definition, the average change of rate is given by:
$AVR = \frac{f(x2)-f(x1)}{x2-x1}$
We will calculate AVR for each of the functions.
We have then:

f(x) = x^2 + 3x interval: [-2, 3]:
$f(-2) = x^2 + 3x = (-2)^2 + 3(-2) = 4 - 6 = -2

f(3) = x^2 + 3x = (3)^2 + 3(3) = 9 + 9 = 18$
$AVR = \frac{-2-18}{-2-3}$
$AVR = \frac{-20}{-5}$
$AVR = 4$

f(x) = 3x - 8 interval: [4, 5]:
$f(4) = 3(4) - 8 = 12 - 8 = 4 f(5) = 3(5) - 8 = 15 - 8 = 7$
$AVR = \frac{7-4}{5-4}$
$AVR = \frac{3}{1}$
$AVR = 3$

f(x) = x^2 - 2x interval: [-3, 4]
$f(-3) = (-3)^2 - 2(-3) = 9 + 6 = 15

f(4) = (4)^2 - 2(4) = 16 - 8 = 8$
$AVR = \frac{8-15}{4+3}$
$AVR = \frac{-7}{7}$
$AVR = -1$

f(x) = x^2 - 5 interval: [-1, 1]
$f(-1) = (-1)^2 - 5 = 1 - 5 = -4

f(1) = (1)^2 - 5 = 1 - 5 = -4$
$AVR = \frac{-4+4}{1+1}$
$AVR = \frac{0}{2}$
$AVR = 0$

Answer:
these functions from the greatest to the least value based on the average rate of change are:
f(x) = x^2 + 3x
f(x) = 3x - 8
f(x) = x^2 - 5
f(x) = x^2 - 2x

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3 years ago

PLEASE HELP!!! :))

Nina [5.8K]

For a better understanding of the solution given here please find the attached file which has the relevant diagram.

To answer this question we will have to make use of <u>the Isosceles Triangle Theorem</u> which states that the perpendicular bisector of the base of an isosceles triangle is also the angle bisector of the vertex angle. Thus, as a corollary we know that is EF is the angle bisector of the vertex angle ∠E, then, EF is the perpendicular bisector of the of the base DK.

Please follow the diagram of a complete understanding of the logic and the solution.

As EF is the angle bisector as given in the question, thus we will have:

$m\angle DEK=2\times m\angle DE F=2\times 43^{\circ}=86^{\circ}$ .

Also, from the Theorem we know that KF will be half of DK and thus, KF will be: $KF=\frac{1}{2}DK= \frac{1}{2}\times 16=8$ centimeters.

Likewise, from the same theorem we have: $m\angle EFD=90^{\circ}$

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3 years ago