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Elena-2011 [213]
3 years ago
10

Lena reflected this figure across the x-axis. She writes the vertices of the image as A'(−2, 8), B'(−5, 6), C'(−8, 8), D'(−4, 2)

. Lena is incorrect because
A) she reflected the pre-image across the y-axis.

B) she reflected the pre-image across the line y = x.

C) she rotated the pre-image 90 degrees counterclockwise.

D) she rotated the pre-image 180 degrees counterclockwise.

Mathematics
1 answer:
garik1379 [7]3 years ago
8 0

Answer:

Option A) she reflected the pre-image across the y-axis

Step-by-step explanation:

we know that

When reflect a point across the the x-axis the rule of the reflection is equal to

(x,y) -----> (x,-y)

When reflect a point across the the y-axis the rule of the reflection is equal to

(x,y) -----> (-x,y)

In this problem we have

A(2,8),B(5,6),C(8,8),D(4,2)

Applying the rule of the reflection across the x-axis

The coordinates of the image would be

A'(2,-8),B'(5,-6),C'(8,-8),D'(4,-2)

Applying the rule of the reflection across the y-axis

The coordinates of the image would be

A'(-2,8),B'(-5,6),C'(-8,8),D'(-4,2)

therefore

she reflected the pre-image across the y-axis.

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A 1.5-mm layer of paint is applied to one side of the following surface. Find the approximate volume of paint needed. Assume tha
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Answer:

V = 63π / 200  m^3

Step-by-step explanation:

Given:

- The function y = f(x) is revolved around the x-axis over the interval [1,6] to form a spherical surface:

                                 y = √(42*x - x^2)

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Find:

The volume of paint needed

Solution:

- Let f be a non-negative function with a continuous first derivative on the interval [1,6]. The Area of surface generated when y = f(x) is revolved around x-axis over the interval [1,6] is:

                           S = 2*\pi \int\limits^a_b { [f(x)*\sqrt{1 + f'(x)^2} }] \, dx

- The derivative of the function f'(x) is as follows:

                            f'(x) = \frac{21-x}{\sqrt{42x-x^2} }

- The square of derivative of f(x) is:

                            f'(x)^2 = \frac{(21-x)^2}{42x-x^2 }

- Now use the surface area formula:

                           S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2} *\sqrt{1 + \frac{(21-x)^2}{42x-x^2 } }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+(21-x)^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+441-42x+x^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{441} }] \, dx\\S = 2*\pi \int\limits^6_1 { 21} \, dx\\\\S = 42*\pi \int\limits^6_1 { dx} \,\\\\S = 42*\pi [ 6 - 1 ]\\\\S = 42*5*\pi \\\\S = 210\pi

- The Volume of the pain coating is:

                           V = S*t

                           V = 210*π*3/2000

                          V = 63π / 200 m^3

8 0
3 years ago
Read 2 more answers
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