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bogdanovich [222]
3 years ago
13

A right rectangular prism is 4 inches long and 3 inches wide and 6 inches tall what are the area and perimeter of a cross sectio

n that is parallel to the base of the prism
Mathematics
2 answers:
Alina [70]3 years ago
8 0

12 and 14 I know because i just did this and got it correct

zzz [600]3 years ago
6 0
The area of a cross section paralell to the base is the base
the base is 4inches by 3 inches
area is 12 square inches
perimiter is 2(L+W)=2(4+3)=2(7)=14 inches


area is 12 square inches
perimiter is 14 inches
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This diagram shows triangles ABC and PQR.
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<h3>A reflection across the line x=3, a reflection across the x-axis and a dilation with a scale factor of 2, because each side is double.</h3><h3>Step-by-step explanation:</h3>

We know that the first transfomration is a rotation 90° clockwise.

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What number would ypu add to both sides of x^2+7x=4 to complete the square
Tcecarenko [31]

For a monic quadratic (factor of 1 on x^2) we always add half the coefficient on the linear (x) term, so

\left(\dfrac 7 2\right)^2 = \dfrac{49}{4}

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6 0
3 years ago
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The pumpkin cost $3.96

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3 years ago
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
Mariulka [41]

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)

- The surface is coated with paint with uniform layer thickness t = 1.5 mm

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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