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amid [387]
3 years ago
9

A cicular pool with a radius of 12 meter if filled with water to a depth of 3 meters. what is the weight of water in the swimmin

g pool?
Mathematics
1 answer:
Tcecarenko [31]3 years ago
5 0
The weight of water in swimming pool is 13305600kg
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Pleaaaaseeeeeee help!!!!!!!!!!!!!!!!!!!!!<br> brainliest and 20 points
beks73 [17]

Answer:

x=0

Step-by-step explanation:

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Shkiper50 [21]

Answer:

infinite

Step-by-step explanation:

  • x=1
  • y=3

Let the linear equation in two variables be ax+by+c=0

Put values

\\ \sf\longmapsto 1a+3b+c=0

\\ \sf\longmapsto a+3b+c

Hence for any values it has infinite number of solutions .

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Divide 82 in the ratio 1:3<br><br> will give brainliest to whoever answers first
Hatshy [7]

Answer:

20.5 : 61.5

Step-by-step explanation:

1:3

1 + 3 = 4

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ratio of 1:3

1 × 20.5 = 20.5

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You previously found the mean of this data set. Use that in answering the question. 63, 89, 92, 73, 79, 72, 34, 36, 94, 21, 25,
kicyunya [14]

Answer:

Sum of squares of differences = 11239.74

Step-by-step explanation:

We are given the following data set:

63, 89, 92, 73, 79, 72, 34, 36, 94, 21, 25, 93, 22, 90, 79

We have to calculate the sum of square of the data set.

Formula:

\text{Sum of square of differences} = \sum (x_i -\bar{x})^2

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{962}{15} = 64.13

Sum of squares of differences =

1.284444444 + 618.3511113 + 776.5511113 + 78.61777778 + 221.0177779 + 61.88444445 + 908.0177776 + 791.4844443 + 892.017778 + 1860.484444 + 1531.417778 + 833.2844446 + 1775.217777 + 669.0844446 + 221.0177779

= 11239.74

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3 years ago
Find the angel x in d diagram above with a well structured solution.<br><br>Thanks​
Tom [10]

Step-by-step answer:

Please refer to attached image.

1. Quad PQRS is cyclic (all vertices on the same circle), so opposite angles are supplementary, meaning

  that

  angles QPS and QRS are supplementary =>

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2. Triangle RSQ is isosceles with RS=RQ =>

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3. Angle RSQ = (180 - 106) /2 = 74 / 2  = 37

4. QP is a diameter => angle QSP is a right-angle.

5. From (3) and (4) above,

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6. Since PQRS is cyclic, angles RQP  and RSP are supplementary =>

   RQP+RSP = 180   =>

   x + 127 = 180   =>

   x = 180 - 127 = 53 degrees.

5 0
3 years ago
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