Question

I need urgent help!! I want someone to help me find examples proving something right in one question, and helping me solve another. I will medal and fan!

Answer 1

Alright, I will give my answer to number 2, because I need help with number 3(I want someone to check my work as I do it) and I know that Dwayne is correct, I just need examples proving that. :)

Answer 2

My answer for number 2: 2. My function is f(x)=500(1.30)^x. The principal rate is 500 and the growth rate is .30, or 30%. The domain is 1<x<20 in years, because time cannot be negative. The range is 500<y<10000, positive integers because you cannot have half a frog, or a negative number of frogs. The principal amount will just indicate if you start with a larger y-value or a smaller one on the graph and the growth rate will ultimately determine the slope or the rate at which y increases from the starting value/principal amount. The domain and range set the boundaries for how high the numbers can get.

Answer 3

So for 3, would I replace 3 and 5 for x, then solve?And one at a time of course.

Answer 4

@Gebooors can you help?

Answer 5

It looks you are managing well!

Answer 6

So, that is correct, what I have for 3? Now, please help me make some examples for 4, I only was just told he's correct when I asked for help, which isn't much help for the full question.

Answer 7

Did you calculate for x = 3 and x= 5?
And you noticed that there are two years when calculating average
Growth?

Answer 8

Quick question, if my problem for number 3 is f(3)=500(1.30)^3, what do I multiply first? What order do I work in?

Answer 9

Calculate first 1.3^3. Then Multiply by 500

Answer 10

Thanks! I will tell you what I get

Answer 11

f(3)=1098.5?

f(3)=500(1.30)^3
f(3)=500(2.297)
f(3)=1098.5

Answer 12

oops 500(2.197)

Answer 13

typed it wrong

Answer 14

Rounded to 1099 is right! Good work!

Answer 15

Yay! Are you sure I round?

Answer 16

You told there are no half frogs :)

Answer 17

oops, you are right! Sorry I missed that step in my problem

Answer 18

How about x= 5? In the same way

Answer 19

Ok, my rounded answer is 1856.

f(5)=500(1.30)^5
f(5)=500(3.71293)
f(5)=1856.456
rounded to 1856

Answer 20

Am I correct?

Answer 21

Right! Good work!

Answer 22

Yay, thanks!

Answer 23

Now you think average between two years?

Answer 24

So I find the average between those and then years 5 and 7?

Answer 25

Let's calculate it for x= 3 and x= 5.

Answer 26

So for years 5 and 3, the average is 1478, rounded? I added them together and divided by 2

Answer 27

X= 3. Y = 1099
X= 5. Y= 1856

You get 1856 = 1099 * q^2

Answer 28

Um, ok?

Answer 29

Sorta makes sense

Answer 30

I think it is question of average growth per cent.

What do you think?

Answer 31

You have to Solve q = square root (1856/1099)

Answer 32

ok

Answer 33

I got q = 1.2995 and average percent is 29,95 also 30 per cent.
We rounded numbers.

But average growth is difference divided by two

Answer 34

It gives how Many frogs comet in average in a year

Answer 35

So finding the average rate of change, which do I do? what should I get as my final answer?

Answer 36

I have to log out now but for the question 4:

I think because lenght is an irrational number and you Multiply height
With irrational number, you get arkea, which is an irrational number.

Answer 37

I believe @phi may help you forward. Good work!

Answer 38

Ok, thanks @Gebooors!!

Answer 39

You're welcome @horsegirl27

Answer 40

I would tweak your answer to 2.)
The principal rate is 500 should read
principal amount is 500
the principal amount is the number of frogs at the start (year 0)

for the domain , I would include year 0. so 0 \( \le \) x \( \le\) 20
notice we should say x can be 0 or bigger
I don't know how you decided x has to be less than (or equal?) to 20.
Maybe argue that after 20 years the frogs are so numerous that they no longer grow at 30% per year, and a new growth function must be used?

anyway, for the range, you should include 500 (so use \( \le\), no tjust < \)
and to get the upper limit, evaluate 500*1.3^20
(remember order of operations... exponents before multiply)

Answer 41

Ok, thanks for the suggestions!

Answer 42

average rate of change means change in y divided by change in x (this is the slope of the line that connects the two points at (3, 1099) and (5, 1856)
this rate is in "frogs per year"

Answer 43

And for x having to be less than 20, I chose 20 as my number for this problem, saying that's what my graph is going up to. But I will include what you said.

Answer 44

Ok, so I made the changes for number 2

Answer 45

now for Q3, find the slope of the line. that represents the average rate of change

Answer 46

Now, what change in y? Should I add, subtract, multiply, or divide? Sorry, I know this but I learned it a while ago and I'm just so confused that I'm not sure

Answer 47

y= 500*1.3^x

(if we plot this on an x-y graph)
at x=3, find y (y is the # of frogs)

Answer 48

I think you already did that, up above
you get the point (3, 1099)

Answer 49

Yeah

Answer 50

So @phi what is my average for years 3 and 5 and 5 and 7? Since I'm not really understanding it when u try to help me do it myself, could you tell me the answers, then explain how you found them? Then, I will still be learning, but not confused

Answer 51

Here is a picture

Answer 52

you should find the number of frogs at x=3 (you get 1099 )
and the number at x=5
now form the (x,y) pairs (the points A and B in the graph)

now find the slope between those two points
slope is change in y divided by change in x

Answer 53

can you do that ?

Answer 54

Yeah let me do it one sec

Answer 55

Ok either 379(rounded up) or 1478(also rounded) I did each a different way because I wasn't sure... Is one of them correct? idk...

Answer 56

379 is good. that means the frog population grows (on average) about 379 frogs per year
between years 3 and 5

Answer 57

now find the (x,y) pair for x= 7
and do the same thing for years 5 and 7

Answer 58

ok

Answer 59

Ok I got 641. For f(7) I got 3137

Answer 60

you should get a faster rate of change because this second line has a steeper slope
your number looks good

Answer 61

ok

Answer 62

the average rate of change is how much the population is growing in "frogs per year"
during the specified interval

Answer 63

ok

Answer 64

for Q4

if you add irrational numbers to another irrational or rational number, you get an irrational number. (the only exception I can think of is , for example, sqr(5) + - sqr(5) = 0 )
that means if one side (which means 2 sides) of a rectangle have irrational length
we expect the perimeter to also be irrational

Answer 65

although thinking about it, it is possible (as an example) for the length to be
10+sqr(2) and this is irrational
and the width to be
5 - sqr(2) and this is irrational
but when we add up the 4 sides to figure out the perimeter:

10+sqr(2) + 5 - sqr(2) + 10 + sqr(2) + 5 - sqr(2)
we will get 10+5+10+5 = 30 (the sqr(2) parts add up to 0)
and 30 is rational

Answer 66

now the area = L*W
if L is sqr(12) and W= sqr(3)
the L*W = \( \sqrt{12} \cdot \sqrt{3}= \sqrt{12\cdot 3} = \sqrt{36}=6\)

Answer 67

ok

Answer 68

Sorry I was gone for the last minute or so back now

Answer 69

in my first example, the perimeter is rational, and the area will be irrational in the second example, perimeter is irrational and the area is rational
for a third example:
L= 10+ sqr(2) W= 10- sqr(2)
here the perimeter will be 40
and the area = \( (10 +\sqrt{2})(10- \sqrt{2}) = 100-2 = 92 \) (fixed my math!)
and both the area and perimeter are rational

Answer 70

although 100-2= 98 not 92

Answer 71

ok

Answer 72

thx

Answer 73

yw