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kogti [31]
3 years ago
5

The ordered pairs (1, 81), (2, 100), (3, 121), (4, 144), and (5, 169) represent a function. What is a rule that represents this

function?
y = x^8

y = 8^x

y = (x + 8)^2

y = 8x ...?
Mathematics
2 answers:
nadezda [96]3 years ago
5 0

Answer:

y=(x+8)^{2}

Step-by-step explanation:

we have that

The ordered pairs (1, 81), (2, 100), (3, 121), (4, 144), and (5, 169) represent a function

The rule that that represent this function is equal to

y=(x+8)^{2}

Verify

For x=1 -----> Find the value of y in the equation

y=(1+8)^{2}=81 -----> is correct

For x=2 -----> Find the value of y in the equation

y=(2+8)^{2}=100 -----> is correct

For x=3 -----> Find the value of y in the equation

y=(3+8)^{2}=121 -----> is correct

For x=4 -----> Find the value of y in the equation

y=(4+8)^{2}=144 -----> is correct

For x=5 -----> Find the value of y in the equation

y=(5+8)^{2}=169 -----> is correct


dezoksy [38]3 years ago
3 0
To determine what function is the correct answer we use the choices listed above to substitute x and obtain the value of y and see if it corresponds to the given y value. We do as follows:

<span>y = x^8
y = (1)^8
y = 1 ----> NOT THE ANSWER

y = 8^x
y = 8^(1)
y = 8</span>---> NOT THE ANSWER<span>

y = (x + 8)^2
y = (1+8)^2
y = 81

y = 8x
y = 8(1)
y = 8 </span>---> NOT THE ANSWER
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trapecia [35]

Answer:

Solution ( Second Attachment ) : - 2.017 + 0.656i

Solution ( First Attachment ) : 16.140 - 5.244i

Step-by-step explanation:

Second Attachment : The quotient of the two expressions would be the following,

6\left[\cos \left(\frac{2\pi }{5}\right)+i\sin \left(\frac{2\pi \:}{5}\right)\right] ÷ 2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]

So if we want to determine this expression in standard complex form, we can first convert it into trigonometric form, then apply trivial identities. Either that, or we can straight away apply the following identities and substitute,

( 1 ) cos(x) = sin(π / 2 - x)

( 2 ) sin(x) = cos(π / 2 - x)

If cos(x) = sin(π / 2 - x), then cos(2π / 5) = sin(π / 2 - 2π / 5) = sin(π / 10). Respectively sin(2π / 5) = cos(π / 2 - 2π / 5) = cos(π / 10). Let's simplify sin(π / 10) and cos(π / 10) with two more identities,

( 1 ) \cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos \left(x\right)}{2}}

( 2 ) \sin \left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos \left(x\right)}{2}}

These two identities makes sin(π / 10) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}, and cos(π / 10) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}.

Therefore cos(2π / 5) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}, and sin(2π / 5) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}. Substitute,

6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right] ÷ 2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]

Remember that cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting those values,

6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right] ÷ 2\sqrt{2}\left[0-i\right]

And now simplify this expression to receive our answer,

6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right] ÷ 2\sqrt{2}\left[0-i\right] = -\frac{3\sqrt{5+\sqrt{5}}}{4}+\frac{3\sqrt{3-\sqrt{5}}}{4}i,

-\frac{3\sqrt{5+\sqrt{5}}}{4} = -2.01749\dots and \:\frac{3\sqrt{3-\sqrt{5}}}{4} = 0.65552\dots

= -2.01749+0.65552i

As you can see our solution is option c. - 2.01749 was rounded to - 2.017, and 0.65552 was rounded to 0.656.

________________________________________

First Attachment : We know from the previous problem that cos(2π / 5) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}, sin(2π / 5) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}, cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting we receive a simplified expression,

6\sqrt{5+\sqrt{5}}-6i\sqrt{3-\sqrt{5}}

We know that 6\sqrt{5+\sqrt{5}} = 16.13996\dots and -\:6\sqrt{3-\sqrt{5}} = -5.24419\dots . Therefore,

Solution : 16.13996 - 5.24419i

Which rounds to about option b.

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Step-by-step explanation:

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