Can you see if im right?
The first is incorrect. You can indeed take the roots of both 64 and 4, not just 4.
we should not take your test for you
8^2 = 64, therefore sqrt(64) = 8.
Is it a test?
You're not. I'm asking if I'm right. And no, my practice before the test.
5 is irrational..
the first is correct
root of prime is irational
5 is not irrational, 5 is prime, and thus sqrt(5) is irrational
1 & 2 are correct 3 is incorrect
:o
the sqr of 5 is
why is 3 wrong :D
D: *
Oh awks, there's an a and b, I didn't see that sorry.
Do you know the formula for compound interest? The formula for depreciation is similar.
No D:
oh, wait is it A
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
What was the rate of depreciation?
That's the thing, I couldn't find the rate :(
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?
yea
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.
Do you remember how to calculate percent change?
Divide?
Of course, being math, there's a formula for that too: \( percent~change = \dfrac{new~value - old~value}{old~value} \times 100 \)
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.
-13% ?
Is it B :D
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?
Well the principle is the beginning value, the rate is -13 and t is years, why i think its B
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
ok, so am i right? ... possibly >.>
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
So its was D -facepalm-
Write 0.87 in for 1 - r. Write 54,205 in for P Leave t as t. Which choice is it?
Correct. You got it.
haha ty! can you look at this one too?
I'm helping someone with another problem. I'll be here soon.
k
U_U
Sorry. It took me longer than I thought.
no worries,
typo
You chose choice 4. That means add equation 1 of system 2 to 3 times equation 2 of system 2.
2 and 3 have the same result
Let's look at it. System 2: 4x - 5y = 2 3x - 8y = 4 What is 3 times equation 2?
9x -24y = 12
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.
so its B? wio says its right?
Now let's look at choice 2. It also states that system 2 and 3 are the same, but with a different reason.
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?
12x -32y = 16
i have to get off its 2 33 am for me D:
what should i do :(
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.
4x - 5y = 2 6x - 16y = 8 Add the two eqs above.
10x -21y = 10
oh it was b O_O
Right. That is what is in choice 2.
thank you!!!!!
wlcm
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