Which of the following is true?
1 answer:
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Remark
There is no short way to do this problem and no obvious way to get the answer other that to solve each part.
Solve
A
Multiply by 2
x + 1.6 = 2(x + 0.1) Remove the brackets
x + 1.6 = 2x + 0.1*2
x + 1.6 = 2x + 0.2 Subtract x from both sides
1.6 = x + 0.2 Subtract 0.2 from both sides
1.6 - 0.2 = x
1.4 = x
Circle A
B
Subtract 2x from both sides.
3x - 2x = 1.4
Circle B
C
Remove the brackets.
4x + 6 = 2x - 6 Add 6 to both sides
4x + 12 = 2x Subtract 4x from both sides.
12 = -2x Divide by - 2
12/-2 = x
x = - 6 Don't circle C
D
I'm going to be very scant in my solution of this. You can fill in the steps.
3x = 4.2
x = 4.2/3
x = 1.4
Circle D
Given that a species of beetles grows 32% every year.
So growth rate is given by
r=32%= 0.32
Given that 100 beetles are released into a field.
So that means initial number of beetles P=100
Now we have to find about how many beetles will there be in 10 years.
To find that we need to setup growth formula which is given by
where A is number of beetles at any year n.
Plug the given values into above formula we get:
now plug n=10 years
Hence answer is approx 1606 beetles will be there after 20 years.
Now we have to find about how many beetles will there be in 20 years.
To find that we plug n=20 years
Hence answer is approx 25791 beetles will be there after 20 years.
Now we have to find time for 100000 beetles so plug A=100000
33.174666862=n
Hence answer is approx 33 years.