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NeX [460]
4 years ago
9

Which statement below is correct for the following set of ordered pairs?

Mathematics
1 answer:
stellarik [79]4 years ago
8 0

We observe the set doesn't have two points with the same x value, so this is a function.  Let's look at the answers.

a The set is a function since each element in the domain has a different element in the range.

No, this is more the idea of one-to-one or injectivity.

b The set is a function since each element in the range has a different element in the domain.

This isn't quite right either but we're going to choose it as the best answer.  It's not that each element in the range has a different element in the domain; that would be onto aka surjectivity.  It's that each element in the set, each point, has a different element in the domain, a different y coordinate.  It's ok if two different x coordinates map to the same y coordinate, which they don't here but they could.

Answer: b

c The set is not a function since each element in the domain has a different element in the range.

Nope, it is a function.

d The set is not a function since each element in the range has a different element in the domain.

Nope, it is a function.

Demand better questions from your teachers!

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a rectangle is constructed with its base on the x-axis and two of its vertices on the parabola y=121-x^2. what are the dimension
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Answer:

the length is \dfrac{22}{\sqrt{3}} and the width is 121-\left(\dfrac{11}{\sqrt{3}}\right)^2=121-\dfrac{121}{3}=\dfrac{242}{3}.

Step-by-step explanation:

Let points A and B be placed on the x-axis. Their coordinates are A(x_0,0)\ (x_0>0) and B(-x_0,0) (because of parabola symmetry). Two other vertices lie on the parabola, then C(-x_0,121-x_0^2) and D(x_0,121-x_0^2). The length of the side AB is 2x_0 and the length of the side AD is 121-x_0^2. Thus, the area of the rectangle ABCD is

A=2x_0\cdot (121-x_0^2)=242x_0-2x_0^3.

Find the derivative A':

A'=242-2\cdot 3x_0^2=242-6x_0^2.

Equate A' to 0:

242-6x_0^2=0,\\ \\x_0^2=\dfrac{121}{3},\\ \\x_0=\dfrac{11}{\sqrt{3}}.

The maximum area of the rectangle is

A_{max}=242\cdot \dfrac{11}{\sqrt{3}}-2\cdot \left(\dfrac{11}{\sqrt{3}}\right)^3=\dfrac{2662}{\sqrt{3}}-\dfrac{2662}{3\sqrt{3}}=\dfrac{5324}{3\sqrt{3}}\ un^2.

The dimensions of the rectangle are:

the length is \dfrac{22}{\sqrt{3}}\ un. and the width is 121-\left(\dfrac{11}{\sqrt{3}}\right)^2=121-\dfrac{121}{3}=\dfrac{242}{3}\ un.

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