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ICE Princess25 [194]
3 years ago
14

A 6m ladder leans against a wall. The bottom of the ladder is 1.3m from the wall at time =0sec and slides away from the wall at

a rate of 0.3m/s.
Find the velocity of the top of the ladder at time =2 (take the direction upwards as positive).


(Use decimal notation. Give your answer to three decimal places.)
Mathematics
1 answer:
icang [17]3 years ago
5 0

Answer:

- 0.100

Step-by-step explanation:

Length of the ladder,  H = 6 m

Distance at the bottom from the wall, B = 1.3 m

Let the distance of top of the ladder from the bottom at the wall is P

Thus,

from  Pythagoras theorem,

B² + P² = H²    .

or

B² + P² = 6²          ..............(1)      [Since length of the ladder remains constant]

at B = 1.3 m

1.3² + P² = 6²

or

P² = 36 - 1.69

or

P² = 34.31

or

P = 5.857

Now,

differentiating (1)

2B(\frac{dB}{dt})+2P(\frac{dP}{dt})=0

at t = 2 seconds

change in B = 0.3 × 2= 0.6 ft

Thus,

at 2 seconds

B = 1.3 + 0.6 = 1.9 m

therefore,

1.9² + P² = 6²

or

P = 5.69 m

on substituting the given values,

2(1.9)(0.3) + 2(5.69) × (\frac{dP}{dt}) = 0

or

1.14 + 11.38 × (\frac{dP}{dt}) = 0

or

11.38 × (\frac{dP}{dt}) = - 1.14

or

(\frac{dP}{dt}) = - 0.100

here, negative sign means that the velocity is in downward direction as upward is positive

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3 years ago
Kristen invests $ 5745 in a bank. The bank 6.5% interest compounded monthly. How long must she leave the money in the bank for i
podryga [215]

Answer:

1. 10.7 years

2. 17.0 years

3. 2nd option

Step-by-step explanation:

Use formula for compounded interest

A=P\cdot \left(1+\dfrac{r}{n}\right)^{nt},

where

A is final value, P is initial value, r is interest rate (as decimal), n is number of periods and t is number of years.

In your case,

1. P=$5745, A=2P=$11490, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then

11490=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},\\ \\2=(1.0054)^{12t},\\ \\12t=\log_{1.0054}2,\\ \\t=\dfrac{1}{12}\log_ {1.0054}2\approx 10.7\ years.

2. P=$5745, A=3P=$17235, n=12 (compounded monthly), r=0.065 (6.5%) and t is unknown. Then


17235=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12t},

3=(1.0054)^{12t},

12t=\log_{1.0054}3,

t=\dfrac{1}{12}\log_{1.0054}3\approx 17.0\ years.

3. <u>1 choice:</u> P=$5745, n=12 (compounded monthly), r=0.065 (6.5%), t=5 years and A is unknown. Then

A=5745\cdot \left(1+\dfrac{0.065}{12}\right)^{12\cdot 5},\\ \\A=5745\cdot (1.0054)^{60}\approx \$7936.39.

<u>2 choice:</u> P=$5745, n=4 (compounded monthly), r=0.0675 (6.75%), t=5 years and A is unknown. Then

A=5745\cdot \left(1+\dfrac{0.0675}{4}\right)^{4\cdot 5},\\ \\A=5745\cdot (1.0169)^{20}\approx \$8032.58.

The best will be 2nd option, because $8032.58>$7936.39

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Answer:

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Now, we are able to calculate the ultimate cost of coating:

If the cost is $10 per cm^2, then we have to find how much it costs to coat 15.3 cm^2.

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Answer:

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