What’s is the absolute value of -6? use the number line to help answer the question.
1 answer:
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Before we start answering the question, let's define the compound interest formula:
Where:
<span>'A'</span> is the amount of money in dollars
'P' is the principal amount of money in dollars
'r' is the interest rate (decimal)
'n' is the number of times interest is compounded per year
't' is the time in years
<span>(A) Find Principal Amount
</span><u /><span><u>Given:</u>
</span>A = 12,000
P = ?
r = 0.08
n = 2 (semiannually)
t = 5
Now we plug our values in and solve:
∴ You would have to deposit $8106.77 in order to have $12,000 in 5 years from now.
(B) Find Principal Amount
Same given values as above, with the exception of 't' which is now 10 instead of 5.
∴ You would have to deposit $5476.64 in order to have $12,000 in 10 years from now.
Hope this helps!
Where:
<span>'A'</span> is the amount of money in dollars
'P' is the principal amount of money in dollars
'r' is the interest rate (decimal)
'n' is the number of times interest is compounded per year
't' is the time in years
<span>(A) Find Principal Amount
</span><u /><span><u>Given:</u>
</span>A = 12,000
P = ?
r = 0.08
n = 2 (semiannually)
t = 5
Now we plug our values in and solve:
∴ You would have to deposit $8106.77 in order to have $12,000 in 5 years from now.
(B) Find Principal Amount
Same given values as above, with the exception of 't' which is now 10 instead of 5.
∴ You would have to deposit $5476.64 in order to have $12,000 in 10 years from now.
Hope this helps!
Problem 1
Answer: year 4
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The break-even point is when the profit is $0. You neither earn money nor lose it.
Plug in p(x) = 0 and solve for x
p(x) = x^3 - 4x^2 + 5x - 20
x^3 - 4x^2 + 5x - 20 = p(x)
x^3 - 4x^2 + 5x - 20 = 0 ... replace p(x) with 0
(x^3 - 4x^2) + (5x - 20) = 0
x^2(x - 4) + 5(x - 4) = 0
(x^2+5)(x - 4) = 0
x^2+5 = 0 or x-4 = 0
The equation x^2+5 = 0 has no real solutions; however x-4 = 0 solves to x = 4.
So plugging x = 4 into p(x) will lead to p(x) = 0. Meaning that the company breaks even at year 4.
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Problem 2
Answer: choice B) between 2.5 and 3.0; between 4.0 and 4.5
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Look at the f(x) column. Specifically we are looking for the times when the numbers change from positive to negative, or vice versa. Somewhere in between this change, y will have to equal 0 at some point (at least once). Note how in row 2 and row 3, we have f(x) = 1.1 change to f(x) = -0.8; so the change is from positive to negative.
This means f(x) = 0 for some x value between x = 2.5 and x = 3.5. Also, the same kind of logic applies for the last two rows of the table as well pointing to another root between x = 4.0 and x = 4.5 (check out the attached images)
