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aksik [14]
3 years ago
6

A boat leaves a dock at 9:00 PM and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15km/h an

d reaches the same dock at 10:00 PM. How many minutes after 9:00 PM were the two boats closest together? (Round your answer to the nearest minute.)
Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
8 0

Answer:22 min

Step-by-step explanation:

we need to find the closest distance between two ship.

let us take the ships meet after time t

now distance travel by first ship at any time t is given by 20t

the position of second ship is 15-15t

let us take distance between them is d

therefore

D^2=\left ( 20t\right )^2+\left ( 15-15t\right )^2

now to get the shortest distance differentiate D w.r.t time

D^2=625t^2-450t+225

\frac{\mathrm{d}D }{\mathrm{d} t}=1250t-450=0

t=0.36hr

thus at t=0.36hr \approx 21.6min\approx 22 min

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Celia has some cakes.
Vadim26 [7]

The carrot cakes are the fraction of the cake given by 1 - \frac{6}{7} multiplied by \frac{1}{3}

  • The fraction of the cakes that are carrot cakes is \underline {\dfrac{1}{21}}

<h3>How are the cakes evaluated as fractions?</h3>

The fraction of the cake that are chocolate cake = \dfrac{2}{3}

The fraction of the remaining cake that cheesecakes = \dfrac{6}{7}

The rest of the cakes = carrot cakes

Required:

The fraction of the cake that  are carrot cakes

Solution:

The fraction of the cakes that are not chocolate cake = 1 - \dfrac{2}{3} = \mathbf{ \dfrac{1}{3}}

The fraction of the remaining cakes that are cheesecake = \dfrac{6}{7} \times \dfrac{1}{3} = \mathbf{ \dfrac{6}{21}}

The fraction of the cake that are carrot cake = \dfrac{1}{3} - \dfrac{6}{7} \times \dfrac{1}{3} = \mathbf{\dfrac{1}{21}}

  • The fraction of the cakes that are carrot cake is \underline{\dfrac{1}{21}}

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5 0
2 years ago
Solve this equation: 3x-1/5-2/9x=124/5
Blababa [14]
C is the correct answer
4 0
4 years ago
Read 2 more answers
Determine the period.
dimulka [17.4K]

Answer:

4

Step-by-step explanation:

The period is the length of one cycle.

By cycle, I mean the smallest piece that can be traced and copied over and over to form the whole graph. If you notice the graph starts at x=0 and the curve makes a complete cycle by x=4. The cycle starts over again at x=4 and ends at x=8. I hope you are seeing the graph looks exactly the same from x=0 to x=4 as x=4 to x=8.

The length of one cycle of this graph is 4.

Please let me know if you have further questions about this question.

Thank you.

8 0
3 years ago
Factor completely.<br><br> 10,000 - d^4 =<br><br> x^10 y^8 z^8 - x^4 y^9 z^2 =
Luden [163]
1. 
(100 + d^2)(d + 10)(d - 10)

Use the rule of splitting perfect squares to get the factors. 

2. 
x^4y^8z^2(x^2z^6 - y)

The first term is the greatest common factor. 
3 0
3 years ago
A container manufacturer plans to make rectangular boxes whose bottom and top measure 4x by 3x. The container must contain 48in.
alex41 [277]

The height of the container that will be able to minimize the cost will be 3.08cm.

<h3>How to calculate the height?</h3>

The volume of the box will be:

= (3x)(4x)h

= 12x²h

From the information given, we are told that the container must contain 48in³. Therefore,

48 = 12x²h

h = 4/x²

The function cost will be:

= 3.50(2)(12x²) + 4.40(14x)h

= 84x² + 61.6x(4/x²)

= 84x² + 246.4/x

We'll use the first derivative. This will be:

dC/dx = 168x - 246.4/x²

x = 1.14.

Therefore, the height will be:

h = 4/x² = 4/1.14² = 3.08cm

In conclusion, the height is 3.08cm.

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4 0
3 years ago
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