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Math Challenge I-C Handout 5

 
 
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Math Challenge I-C Handout 5
by Christina Peng - Tuesday, 10 October 2017, 8:40 PM
 

Could someone help me with these problems? Specifically, I don't understand the question for 5.15 and I don't know how to do 5.28 and 5.30.


Question 5.15:

The domain of the parabola y+3=2(x4)2y+3=−2(x−4)2 is all yy such that yLy≤L for some integer LL. What is LL?

Question 5.28:

Rewrite each of the following functions in the form y=axy=ax for a rational number a>0

Part a): y=122xy=122x.

Part b): y=22x


Question 5.30:

Nickel-63 has a half-life of about 11 century. This means that after every century, half of the nickel-63 you have will decay. Suppose you start out with 22 kg of nickel-63. Write an equation that gives the amount (in kg) of nickel remaining after xxcenturies.

y=Thanks
 
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Re: Math Challenge I-C Handout 5
by John Lensmire - Wednesday, 11 October 2017, 3:00 PM
 

Question 5.15 should be the "range" not the domain. This is now fixed. Thanks for the heads up!

Hint for 5.28:  The key here is going to be using our rules of exponents to combine the numbers. So for part a), we want to first write $\displaystyle \frac{1}{2}$ as $2^\text{something}$ and then combine it with the $2^x$. For part b), take a look at the rules of exponents again. Which rule allows us to say, for example, $(2^3)^4 = 2^{3\times 4} = 2^{12}$?

Hint for 5.30:  Start by making a table answering how much nickel-63 is left after $0$ centuries, after $1$ century, after $2$ centuries, etc. To get started we have $2$ kg after $0$ years and $\frac{1}{2}\times 2 = 1$ kg after $1$ century. Can you find a pattern and then turn this pattern into a formula?

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Re: Math Challenge I-C Handout 5
by Christina Peng - Wednesday, 11 October 2017, 7:53 PM
 

Question 5.28:

Part b): y=22x
y=(2^x)^-2


Thank you! Is this what you mean? I feel like part b is already simplified to that form and my answer is not exactly writing in that form.

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Re: Math Challenge I-C Handout 5
by John Lensmire - Thursday, 12 October 2017, 9:39 AM
 

Almost!

To review what you did, you wrote $2^{-2x} = 2^{(x)\times (-2)} = (2^x)^{-2}$. Is there another way to think about the multiplication of $(x)\times (-2)$ so that you end up with a number raised to the $x$?

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Re: Math Challenge I-C Handout 5
by Christina Peng - Thursday, 12 October 2017, 4:28 PM
 

This is kind of confusing...

Is this what you mean?

Question 5.28:

Part a): y=122xy=122x.

y= (2^-1)(2^x)

y=2^(x-1)

Part b): y=22x

y=(2^-2)^x

y=(1/4)^x


Does the answer have to be like number to the x's power?