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Aleks04 [339]
3 years ago
10

How does the volume of a cylinder change when its diameter is halved? Explain Please.

Mathematics
1 answer:
NNADVOKAT [17]3 years ago
8 0
The answer is: it will decrease 4 times.

Let d be diameter and h height of a cylinder

The volume of a cylinder is:
V1 = π * (d/2)² * h = π * d²/4 * h = π * d² * h / 4
If diameter is halved, the volume is:
V2 = π * ((d/2)/2)² * h = π * (d/4)² * h = π * d²/16 * h = π * d² * h / 16

Now, let's compare them:
V1 / V2 = (π * d² * h / 4) / (π * d² * h / 16)

π * d² * h can be cancelled out:
V1 / V2 = (1/4) / (1/16) = (1/4) * 16 = 16/4 = 4

Therefore, the volume of the smaller cylinder will decrease 4 times.
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Explanation:
Given: 2x^4 - 8x^4
Take the common out:
=> x^4(2 - 8)

Hence: => [tex]-6x^4[/tex] (Option B)

2. Ans: Option (D) -3y^5

Explanation:
Given: 6y^5 - 9y^5
Take the common term(s) out:
=> y^5(6-9)

Hence: => -3y^5 (Option D)

3. Ans: Option (B) 9x^2 - 12x + 6<span> quadratic trinomial 

Explanation:
</span>Given: 6 - 12x + 13x^2 - 4x^2

The standard form of polynomial function must have the highest powered value at the start, then the second highest and so on.

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13x^2 - 4x^2 - 12x + 6
=> 9x^2 - 12x + 6 (Option B)

4. Ans: Option (C) 7.3x^2 + 0.8x + 1.3

Explanation:
In order to find the total number of C<span>ommon and Endler's guppies, you need to add both Common and Endler's guppies' polynomials, as follows:

</span>Common guppies: 3.1x^2 +6.0x + 0.3<span>
Endler's guppies:   </span>4.2x^2 - 5.2x + 1.0
(add both)
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Total number:          7.3x^2 +0.8x + 1.3
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Hence the ans is Option(C) 7.3x^2 + 0.8x + 1.3

5. Ans: Option (D) 2x^2(133 - 18 \pi )

Explanation:
First let's find the total area of the yard:
Total Area of the Yard = 14x * 19x = 266x^2

Now the area of the circular fountain:
Area of the Circular Fountain = \pi r^2
Since, r=6x
Therefore,
Area of the Circular Fountain = \pi (6x)^2 = 36 \pi x^2

Now the final Area of the yard would be:
Final area of the Yard = Total Area of the Yard - Area of the Circular Fountain
Final area of the Yard = 266x^2 - 36 \pi x^2
=> Final area of the Yard = 2x^2(133 - 18 \pi) (Option D)

6. Ans: Option (A) 56x^2

Explanation:
First let's find the total area of the lot:
Total Area of the lot= 6x * 10x = 60x^2

Now the area of the stadium:
Area of the stadium = 1x * 4x = 4x^2

Now the final Area of the lot would be:
Final area of the lot= Total Area of the lot - Area of the stadium
Final area of the lot= 60x^2 - 4x^2
=> Final area of the lot = 56x^2 (Option A)

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