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docker41 [41]
3 years ago
8

in a sponsored walk for charity, 560 students joined. only 0.72% finished the walk. how many students completed the walk?

Mathematics
1 answer:
Keith_Richards [23]3 years ago
7 0

Answer:

About 4 students completed the walk.

Step-by-step explanation:

0.72%=0.0072

0.0072*560=4.032

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Determine the values of k and p so that the solution is "All Reals" when solving for y:
Angelina_Jolie [31]

Answer:

The value of k is  \pm 9 and The value of p is - 23  

Step-by-step explanation:

Given equation as :

81 y - 17  = k² y + 6 + p

The equation has real solution ,

Now for real solution the equation have no solution

∵ The equation has no solution the ,

The coefficient of y must be equal

I.e 81 = k²

Or, k = \sqrt{81}

∴   k = \pm 9

Again , If the coefficient of y is same then the equation is written as

6 + p = - 17

or, p = - 17 - 6

or, p = - 23

Hence The value of k is  \pm 9 and The value of p is - 23   Answer

6 0
3 years ago
Subtract the following complex numbers: (4+4i)-(13+17i) ??
Lady bird [3.3K]

Before combining the real parts and the imaginary parts, line them up vertically as follows:

 4   +   4i

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--------------            Now add up each column.

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6 0
3 years ago
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If a triangle has a height of 12 inches and a base of 5 inches, what's its area? A. 90 sq. in. B. 45 sq. in. C. 60 sq. in. D. 30
Vilka [71]

Answer:

D

Step-by-step explanation:

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4 0
4 years ago
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Answer:

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The numbers of teams remaining in each round of a single-elimination tennis tournament represent a geometric sequence where an i
Anit [1.1K]

Answer:

a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}

Step-by-step explanation:

We are given the following in the question:

The numbers of teams remaining in each round follows a geometric sequence.

Let a be the first the of the geometric sequence and r be the common ration.

The n^{th} term of geometric sequence is given by:

a_n = ar^{n-1}

a_4 = 16 = ar^3\\a_6 = 4 = ar^5

Dividing the two equations, we get,

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the first term can be calculated as:

16=a(\dfrac{1}{2})^3\\\\a = 16\times 6\\a = 128

Thus, the required geometric sequence is

a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}

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3 years ago
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