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kobusy [5.1K]
3 years ago
6

1 1/7 x 3/5 = How do you do this

Mathematics
1 answer:
lidiya [134]3 years ago
6 0

Answer

76/35 or 2 6/35

Step-by-step explanation:

11

7

+

3

5

=  

11 × 5

7 × 5

+

3 × 7

5 × 7

=  

55

35

+

21

35

=  

55 + 21

35

=  

76

35

=  

2 6

35

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The depth of the water at the end of a pier changes periodically along with the movement of tides. On a particular day, low tide
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Draw a diagram to illustrate the problem as shown in the figure below.

The minimum depth of 2.5 m occurs at 12:00 am and at 12:30 pm.
Therefore the period i0s T= 12.5 hours.
The maximum depth of 5.5 m occurs at 6:15 am and at 6:45 pm. Therefore the period of T = 12.5 hours is confirmed.

The double amplitude is 5.5 - 2.5 = 3 m, therefore the amplitude is a = 1.5 m.
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The model for tide depth is
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That is,
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where 
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A plot of the function confirms that the model is correct.

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3 years ago
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Find the vertices and foci of the hyperbola with equation quantity x plus one squared divided by sixteen minus the quantity of y
katrin2010 [14]

Answer:

The vertices are (3 , -5) , (-5 , -5)

The foci are (4 , -5) , (-6 , -5)

Step-by-step explanation:

* Lets study the equation of the hyperbola

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  center (h , k) and transverse axis parallel to the x-axis is

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- The length of the transverse axis is 2 a

- The coordinates of the vertices are  (h  ±  a  ,  k)

- The coordinates of the foci are (h ± c , k), where c² = a² + b²

- The distance between the foci is  2c

* Now lets solve the problem

- The equation of the hyperbola is (x + 1)²/16 - (y + 5)²/9 = 1

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# k = -5

∵ The vertices are (h + a , k) , (h - a , k)

∴ The vertices are (-1 + 4 , -5) , (-1 - 4 , -5)

* The vertices are (3 , -5) , (-5 , -5)

∵ c² = a² + b²

∴ c² = 16 + 9 = 25

∴ c = ± 5

∵ The foci are (h ± c , k)

∴ The foci are (-1 + 5 , -5) , (-1 - 5 , -5)

* The foci are (4 , -5) , (-6 , -5)

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Answer:

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