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Mathematics 22 Online
OpenStudy (anonymous):

Can you see if im right?

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (anonymous):

The first is incorrect. You can indeed take the roots of both 64 and 4, not just 4.

OpenStudy (zzr0ck3r):

we should not take your test for you

OpenStudy (anonymous):

8^2 = 64, therefore sqrt(64) = 8.

OpenStudy (anonymous):

Is it a test?

OpenStudy (anonymous):

You're not. I'm asking if I'm right. And no, my practice before the test.

OpenStudy (anonymous):

5 is irrational..

OpenStudy (zzr0ck3r):

the first is correct

OpenStudy (zzr0ck3r):

root of prime is irational

OpenStudy (zzr0ck3r):

5 is not irrational, 5 is prime, and thus sqrt(5) is irrational

OpenStudy (mathstudent55):

1 & 2 are correct 3 is incorrect

OpenStudy (anonymous):

:o

OpenStudy (anonymous):

the sqr of 5 is

OpenStudy (anonymous):

why is 3 wrong :D

OpenStudy (anonymous):

D: *

OpenStudy (anonymous):

Oh awks, there's an a and b, I didn't see that sorry.

OpenStudy (mathstudent55):

Do you know the formula for compound interest? The formula for depreciation is similar.

OpenStudy (anonymous):

No D:

OpenStudy (anonymous):

oh, wait is it A

OpenStudy (mathstudent55):

Depreciation formula: \(F = P(1 - r)^t\) where F = future value P = present value r = rate of depreciation expressed as a decimal t = number of years

OpenStudy (mathstudent55):

What was the rate of depreciation?

OpenStudy (anonymous):

That's the thing, I couldn't find the rate :(

OpenStudy (mathstudent55):

You can calculate it. The rate of depreciation is constant according to the problem. That means, the percent depreciation from new to year 1 is the same as the rate of depreciation from year 1 to year 2. Ok so far?

OpenStudy (anonymous):

yea

OpenStudy (mathstudent55):

Since the rate of depreciation is constant, all you need to do is find the percent change in value from year 1 to new or from year 2 to year 1. Once again, you only need to do one of these calculations since the rate of depreciation is constant.

OpenStudy (mathstudent55):

Do you remember how to calculate percent change?

OpenStudy (anonymous):

Divide?

OpenStudy (mathstudent55):

Of course, being math, there's a formula for that too: \( percent~change = \dfrac{new~value - old~value}{old~value} \times 100 \)

OpenStudy (mathstudent55):

Let's use the year 1 value and the new value. Then new value = $47,158.35 old value = $54,205 With the formula above and these values you can calculate percent change. This is the same as the depreciation rate.

OpenStudy (anonymous):

-13% ?

OpenStudy (anonymous):

Is it B :D

OpenStudy (mathstudent55):

Great. The percent change was -13%, or a depreciation rate of 13% per year. 13% = 0.13 Now use the formula for depreciation with the original cost as present value, and 0.13 as r. Leave t as t. What do you get?

OpenStudy (anonymous):

Well the principle is the beginning value, the rate is -13 and t is years, why i think its B

OpenStudy (mathstudent55):

Don't use -13%. Use 13%. The formula is made for depreciation, so you need to use the depreciation as a positive amount. 13% as a decimal is 0.13

OpenStudy (anonymous):

ok, so am i right? ... possibly >.>

OpenStudy (mathstudent55):

No. Start with the formula. Replace r with 0.13. What is 1 - 0.13? \(F = P(\color{red}{1 - r})^t\) \(\color{red}{1 - 0.13 = 0.87}\) P = 54,205

OpenStudy (anonymous):

So its was D -facepalm-

OpenStudy (mathstudent55):

Write 0.87 in for 1 - r. Write 54,205 in for P Leave t as t. Which choice is it?

OpenStudy (mathstudent55):

Correct. You got it.

OpenStudy (anonymous):

haha ty! can you look at this one too?

OpenStudy (anonymous):

OpenStudy (mathstudent55):

I'm helping someone with another problem. I'll be here soon.

OpenStudy (anonymous):

k

OpenStudy (anonymous):

U_U

OpenStudy (mathstudent55):

Sorry. It took me longer than I thought.

OpenStudy (anonymous):

no worries,

OpenStudy (anonymous):

OpenStudy (anonymous):

typo

OpenStudy (mathstudent55):

You chose choice 4. That means add equation 1 of system 2 to 3 times equation 2 of system 2.

OpenStudy (anonymous):

2 and 3 have the same result

OpenStudy (mathstudent55):

Let's look at it. System 2: 4x - 5y = 2 3x - 8y = 4 What is 3 times equation 2?

OpenStudy (anonymous):

9x -24y = 12

OpenStudy (mathstudent55):

good Here is eq 1 and 3 times eq 2: 4x - 5y = 2 9x - 24y = 12 Now we add them: 13x - 29y = 14 That is not the second equation of system 3, so choice 4 can't be the answer.

OpenStudy (anonymous):

so its B? wio says its right?

OpenStudy (mathstudent55):

Now let's look at choice 2. It also states that system 2 and 3 are the same, but with a different reason.

OpenStudy (mathstudent55):

Let's take the first equation of system 2: 4x - 5y = 2 Now let's take 2 times the second equation of system 2: 6x - 16y = 8 Now add the two equations together. What do you get?

OpenStudy (anonymous):

12x -32y = 16

OpenStudy (anonymous):

i have to get off its 2 33 am for me D:

OpenStudy (anonymous):

what should i do :(

OpenStudy (mathstudent55):

No. I already multiplied the second eq by 2, all you need to do is add them together. Don't multiply the second eq by 2.

OpenStudy (mathstudent55):

4x - 5y = 2 6x - 16y = 8 Add the two eqs above.

OpenStudy (anonymous):

10x -21y = 10

OpenStudy (anonymous):

oh it was b O_O

OpenStudy (mathstudent55):

Right. That is what is in choice 2.

OpenStudy (anonymous):

thank you!!!!!

OpenStudy (mathstudent55):

wlcm

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