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muminat
3 years ago
14

A cylindrical bucket having inner height and radius 32cm and 8cm respectively, is filled with sand. This bucket is emptied on a

level ground and a conical heap of sand is formed. If the height of the conical heap is 24cm, find the radius and slant height of the heap.​
Mathematics
1 answer:
slavikrds [6]3 years ago
5 0

Answer:

As we know if a solid object turn into another or if a thing which take a shape of one object turn into another object then there volume will be equal.

Take pie = ¶

volume of cylinder = volume of cone

¶×rsquare×h= 1/3¶r square h

18×18×32×3/24=r square

18×18×8×3/6= r square

18×18×4×3/3=r square

√18×18×2×2=r

18×2=r

r =36 cm

h= 24 cm

slant height = √36×36+24×24

slant height = √1872

slant height=

√2×2×2×2×13×3×3

slant height = 12√13 cm

Step-by-step explanation:

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What is the volume of a block of ice measuring 3 meters long, 1.5 meters wide, and 2 meters high? A. 900 cu. m B. 12 cu. m C. 9
lutik1710 [3]
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Read 2 more answers
The difference of two times a number and five is eight more than the number
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part 3. Find the value of the trig function indicated, use Pythagorean theorem to find the third side if you need it.​
Nat2105 [25]

Answer:  \bold{9)\ \sin \theta=\dfrac{1}{3}\qquad 10)\ \sin \theta = \dfrac{4}{5}\qquad 11)\ \cos \theta = \dfrac{\sqrt{11}}{6}\qquad 12)\ \tan \theta = \dfrac{17\sqrt2}{26}}

<u>Step-by-step explanation:</u>

Pythagorean Theorem is: a² + b² = c²   , <em>where "c" is the hypotenuse</em>

<em />

9)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{4}{12}\quad \rightarrow \large\boxed{\dfrac{1}{3}}

Note: 4² + (8√2)² = hypotenuse²   →   hypotenuse = 12

10)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{16}{20}\quad \rightarrow \large\boxed{\dfrac{4}{5}}

Note: 12² + opposite² = 20²   →   opposite = 16

11)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{\sqrt{11}}{6}\quad =\large\boxed{\dfrac{\sqrt{11}}{6}}

Note: adjacent² + 5² = 6²   →   adjacent = √11

12)\ \tan \theta=\dfrac{\text{side opposite of}\ \theta}{\text{side adjacent to}\ \theta}=\dfrac{17}{13\sqrt2}\quad =\large\boxed{\dfrac{17\sqrt2}{26}}

Note: adjacent² + 7² = (13√2)²   →   adjacent = 17

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3 years ago
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