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Determine the ratio of the linear dimensions (e.g., radii or heights) of the similar cylinders.

Mathematics

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: Similar Cylinders

The volumes of two similar cylinders, A and B, are VA=125 cm3V_A = 125 \text{ cm}^3 and VB=27 cm3V_B = 27 \text{ cm}^3 respectively. We need to calculate the area of cylinder B.

Step 1: Determine the ratio of the linear dimensions (e.g., radii or heights) of the similar cylinders.
For similar solids, the ratio of their volumes is equal to the cube of the ratio of their corresponding linear dimensions.
Let LAL_A and LBL_B be corresponding linear dimensions of cylinder A and B.

VAVB=(LALB)3\frac{V_A}{V_B} = \left(\frac{L_A}{L_B}\right)^3

Substitute the given volumes:

125 cm327 cm3=(LALB)3\frac{125 \text{ cm}^3}{27 \text{ cm}^3} = \left(\frac{L_A}{L_B}\right)^3 12527=(LALB)3\frac{125}{27} = \left(\frac{L_A}{L_B}\right)^3

Take the cube root of both sides to find the ratio of linear dimensions:

LALB=125273\frac{L_A}{L_B} = \sqrt[3]{\frac{125}{27}} LALB=53\frac{L_A}{L_B} = \frac{5}{3}

Step 2: Determine the ratio of the surface areas of the similar cylinders.
For similar solids, the ratio of their surface areas is equal to the square of the ratio of their corresponding linear dimensions.
Let AAA_A and ABA_B be the surface areas of cylinder A and B.

AAAB=(LALB)2\frac{A_A}{A_B} = \left(\frac{L_A}{L_B}\right)^2

Substitute the ratio of linear dimensions found in Step 1:

AAAB=(53)2\frac{A_A}{A_B} = \left(\frac{5}{3}\right)^2 AAAB=259\frac{A_A}{A_B} = \frac{25}{9}

Step 3: Conclude regarding the area of cylinder B.
From the ratio of areas, we have AB=925AAA_B = \frac{9}{25} A_A.
The problem asks to "Calculate the area of cylinder B". However, the area of cylinder A (AAA_A) is not provided, nor are any specific dimensions (like radius or height) that would allow us to calculate AAA_A. Therefore, the actual numerical value for the area of cylinder B cannot be determined from the given information alone. We can only express it in terms of the area of cylinder A.

The area of cylinder B is not determinable without the area of cylinder A\boxed{\text{not determinable without the area of cylinder A}}.


: Parallel Lines

Given that y=2xx+6y = 2^x x + 6 and 4+2y=16x4 + 2y = 16x are two parallel lines, we need to find the value of xx.

Step 1: Find the slope of the second line.
The second equation is 4+2y=16x4 + 2y = 16x.
To find its slope, rearrange it into the slope-intercept form y=mx+cy = mx + c, where mm is the slope.

2y=16x−42y = 16x - 4 y=16x−42y = \frac{16x - 4}{2}
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