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Sati [7]
3 years ago
5

The domain and range is?

Mathematics
1 answer:
Molodets [167]3 years ago
3 0

Answer:

In inequality notation:

Domain: -1 ≤ x ≤ 3

Range: -4 ≤ x ≤ 0

In set-builder notation:

Domain: {x | -1 ≤ x ≤ 3 }

Range: {y | -4 ≤ x ≤ 0 }

In interval notation:

Domain: [-1, 3]

Range: [-4, 0]

Step-by-step explanation:

The domain is all the x-values of a relation.

The range is all the y-values of a relation.

In this example, we have an equation of a circle.

To find the domain of a relation, think about all the x-values the relation can be. In this example, the x-values of the relation start at the -1 line and end at the 3 line. The same can be said for the range, for the y-values of the relation start at the -4 line and end at the 0 line.

But what should our notation be? There are three ways to notate domain and range.

Inequality notation is the first notation you learn when dealing with problems like these. You would use an inequality to describe the values of x and y.

In inequality notation:

Domain: -1 ≤ x ≤ 3

Range: -4 ≤ x ≤ 0

Set-builder notation is VERY similar to inequality notation except for the fact that it has brackets and the variable in question.

In set-builder notation:

Domain: {x | -1 ≤ x ≤ 3 }

Range: {y | -4 ≤ x ≤ 0 }

Interval notation is another way of identifying domain and range. It is the idea of using the number lines of the inequalities of the domain and range, just in algebriac form. Note that [ and ] represent ≤ and ≥, while ( and ) represent < and >.

In interval notation:

Domain: [-1, 3]

Range: [-4, 0]

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Answer:

a) Similar

b) Perimeter of rectangle 2 is k times the Perimeter of rectangle 1 (Proved Below)

c) Area of rectangle 2 is k^2 times the Area of rectangle 1 (Proved Below)

Step-by-step explanation:

Given:

Rectangle 1 has length = x

Rectangle 1 has width = y

Rectangle 2 has length = kx

Rectangle 2 has width = ky

(a) Are Rectangle 1 and Rectangle 2 similar? Why or why not?

Rectangle 1 and Rectangle 2 are similar because the angles of both rectangles are 90° and the sides of Rectangle 2 is k times the sides of Rectangle 1. So sides of both rectangles is equal to the ratio k.

(b) Write a paragraph proof to show that the perimeter of Rectangle 2 is k times the perimeter of Rectangle 1.

Perimeter of Rectangle = 2*(Length + Width)

Perimeter of Rectangle 1 = 2*(x+y) = 2x+2y

Perimeter of Rectangle 2 = 2*(kx+ky) = 2kx + 2ky

                                          = k(2x+2y)

                                          = k(Perimeter of Rectangle 1)

Hence proved that Perimeter of rectangle 2 is k times the perimeter of rectangle 1.

(c) Write a paragraph proof to show that the area of Rectangle 2 is k^2 times the area of Rectangle 1.

Area of Rectangle = Length * width

Area of Rectangle 1 = x * y

Area of Rectangle 2 = kx*ky

                                  = k^2 (xy)

                                  = k^2 (Area of rectangle 1)

Hence proved that area of rectangle 2 is k^2 times the area of rectangle 1.

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The Volume of the given solid using polar coordinate is:\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

V= \frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

<h3>What is Volume of Solid in polar coordinates?</h3>

To find the volume in polar coordinates bounded above by a surface z=f(r,θ) over a region on the xy-plane, use a double integral in polar coordinates.

Consider the cylinder,x^{2}+y^{2} =1 and the ellipsoid, 4x^{2}+ 4y^{2} + z^{2} =64

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So, the volume of the solid is given by:

V= \int\limits^{2\pi}_ 0 \int\limits^1_0{} \, [\sqrt{64-4r^{2} }- (-\sqrt{64-4r^{2} })] r dr d\theta

= 2\int\limits^{2\pi}_ 0 \int\limits^1_0 \, r\sqrt{64-4r^{2} } r dr d\theta

To solve the integral take, 64-4r^{2} = t

dt= -8rdr

rdr = \frac{-1}{8} dt

So, the integral  \int\ r\sqrt{64-4r^{2} } rdr become

=\int\ \sqrt{t } \frac{-1}{8} dt

= \frac{-1}{12} t^{3/2}

=\frac{-1}{12} (64-4r^{2}) ^{3/2}

so on applying the limit, the volume becomes

V= 2\int\limits^{2\pi}_ {0} \int\limits^1_0{} \, \frac{-1}{12} (64-4r^{2}) ^{3/2} d\theta

=\frac{-1}{6} \int\limits^{2\pi}_ {0} [(64-4(1)^{2}) ^{3/2} \; -(64-4(2)^{0}) ^{3/2} ] d\theta

V = \frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

Since, further the integral isn't having any term of \theta.

we will end here.

The Volume of the given solid using polar coordinate is:\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta

Learn more about Volume in polar coordinate here:

brainly.com/question/25172004

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