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Artemon [7]
3 years ago
13

The radius of a sphere and of a cylinder are the same. The diameter of the sphere and the height of the cylinder are also the sa

me and are twice the length of the radius. If two cones are formed within the cylinder, as shown in the diagram, then the volume of the sphere is equal to which of the following?
(Points : 1)
the volume of the cylinder + 2(volume of one cone)

the volume of the cylinder 2(volume of one cone)

2(volume of one cone)

the volume of one cone

Mathematics
1 answer:
emmasim [6.3K]3 years ago
4 0
First we have to know the formula of the volume f each of the solids, 

<span>V of sphere = 4/3 pi r^3
</span><span>Volume of Cylinder = pi r^2(2r)=2pi r^3
</span><span>Volume of cone = 1/3 pi r^2(r)=1/3 pi r^3
</span>
The surest and easiest way we can answer this is actually assigning values. We first assign values to r hence we would get the volume of the sphere and rest of the solids (cylinder and cone). You then compare your answers to that of the sphere, and you should get your answer. 
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Which situation is best represented by the equation? 35p + 129.50 = 479.50
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A

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Please help me on this problem
Anna71 [15]

Answer:

The pairs of integer having two real solution forax^{2} -6x+c = 0 are

  1. a = -4, c = 5
  2. a = 1, c = 6
  3. a = 2, c = 3
  4. a = 3, c = 3

Step-by-step explanation:

Given

ax^{2} -6x+c = 0

Now we will solve the equation by putting all the 6 pairs so we get the  following

-3x^{2} -6x-5 = 0 for a = -3 , c=-5

-4x^{2} -6x+5 = 0 for a = -4 , c=5

1x^{2} -6x+6 = 0 for a = 1 , c=6

2x^{2} -6x+3 = 0 for a = 2 , c=3

3x^{2} -6x+3 = 0 for a = 3 , c=3

5x^{2} -6x+4 = 0 for a = 5 , c=4

The above  all are Quadratic equations inn general form ax^{2} +bx+c=0

where we have a,b and c constant values

So for a real Solution we must have

Disciminant , b^{2} -4\timesa\timesc \geq 0

for a = -3 , c=-5 we have

Discriminant =-24 which is less than 0 ∴ not a real solution.

for a = -4 , c=5 we have

Discriminant = 116 which is greater than 0 ∴ a real solution.

for a = 1 , c=6 we have

Discriminant =12 which is greater than 0 ∴ a real solution.

for a = 2 , c=3 we have

Discriminant =12 which is greater than 0 ∴ a real solution.

for a = 3 , c=3 we have

Discriminant =0 which is equal to 0 ∴ a real solution.

for a = 5 , c=4 we have

Discriminant =-44 which is less than 0 ∴ not a real solution.

7 0
3 years ago
In the figure, TP and TS are opposite rays. TQ bisects &lt; RTP.
Ket [755]

Answer:

search-icon-image

Class 9

>>Maths

>>Quadrilaterals

>>Quadrilaterals and Their Various Types

>>In Fig. 6.43, if PQ PS, PQ∥ SR, SQR = 2

Question

Bookmark

In Fig. 6.43, if PQ⊥PS,PQ∥SR,∠SQR=28

0

and ∠QRT=65

0

, then find the values of x and y.

463685

expand

Medium

Solution

verified

Verified by Toppr

Given, PQ⊥PS,PQ∥SR,∠SQR=28

∘

,∠QRT=65

∘

According to the question,

x+∠SQR=∠QRT (Alternate angles as QR is transversal.)

⇒x+28

∘

=65

∘

⇒x=37

∘

Also ∠QSR=x

⇒∠QSR=37

∘

Also ∠QRS+∠QRT=180

∘

(Linear pair)

⇒∠QRS+65

∘

=180

∘

⇒∠QRS=115

∘

Now, ∠P+∠Q+∠R+∠S=360

∘

(Sum of the angles in a quadrilateral.)

⇒90

∘

+65

∘

+115

∘

+∠S=360

∘

⇒270

∘

+y+∠QSR=360

∘

⇒270

∘

+y+37

∘

=360

∘

⇒307

∘

+y=360

∘

⇒y=53

∘

Step-by-step explanation:

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2 years ago
Simplify 3x^2y^4 x 2x^6y
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Answer:

6x^8y^5

Step-by-step explanation:

(3)(2)= 6

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