On mars if you hit a baseball, the height of the ball
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You can do this using synthetic division, which is the easiest way. If x - 2 = 0, then x = 2. That 2 will go outside the "box" and the leading coefficients of the terms in the polynomial will go inside the "box". 2 (1 -3 -10 24). Bring down the first number, the 1. Multiply that 1 by the 2 to get 2. Put that 2 up under the -3 and add to get -1. Multiply that -1 by the 2 to get -2. Put that =-2 up under the -10 and add to get -12. Multiply that -12 by the 2 to get -24. Put the -24 up under the 24 and add to get 0. That means that x - 2 is a factor of the polynomial. What's left, the bolded numbers, are the coefficients of a new polynomial that is one degree less than the polynomial you started with. In other words, when we divide your polynomial by x-2, you get 
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<span>The correct answer is $1452.50 per year.
Explanation:<span>
We use the formula
</span></span>![P(1+\frac{r}{n})^{nt}+PMT(\frac{[(1+\frac{r}{n})^{nt}-1]}{\frac{r}{n}})\times (1+\frac{r}{n})](https://tex.z-dn.net/?f=P%281%2B%5Cfrac%7Br%7D%7Bn%7D%29%5E%7Bnt%7D%2BPMT%28%5Cfrac%7B%5B%281%2B%5Cfrac%7Br%7D%7Bn%7D%29%5E%7Bnt%7D-1%5D%7D%7B%5Cfrac%7Br%7D%7Bn%7D%7D%29%5Ctimes%20%281%2B%5Cfrac%7Br%7D%7Bn%7D%29)
,<span><span>
where P is the amount of principal invested, r is the interest rate as a decimal number, n is the number of times per year the interest is compounded, PMT is the monthly deposit added, and t is the number of years.
Since the amount of principal is not stated, we will assume that Bob is depositing the same amount every following year as he does the first year, so we will let PMT=P.
Our interest rate, r, is 4.2%; 4.2%=4.2/100=0.042.
The number of times the interest is compounded annually, n, is 1.
The amount of time, t, is 7.
We know he wants $50,000. This gives us the equation
50000=P(1+0.042/1)</span></span>⁽¹ˣ⁷⁾<span><span>+P{[(1+0.042/1)</span></span>⁽¹ˣ⁷⁾<span><span>]/(0.042/1)}*(1+0.042/1).
Simplifying this a bit, we have
50000=P(1.042)</span></span>⁷<span><span>+P((1.042</span></span>⁷<span><span>)/0.042)*(1.042).
We can factor out P, giving us
50000=P[1.042</span></span>⁷<span><span>+((1.042</span></span>⁷<span><span>)/0.042)*1.042].
This then gives us
50000=P(34.4234).
Divide both sides:
50000/34.4234 = (P(34.4234))/34.4234,
which gives us P=1452.50.</span></span>
Explanation:<span>
We use the formula
</span></span>
where P is the amount of principal invested, r is the interest rate as a decimal number, n is the number of times per year the interest is compounded, PMT is the monthly deposit added, and t is the number of years.
Since the amount of principal is not stated, we will assume that Bob is depositing the same amount every following year as he does the first year, so we will let PMT=P.
Our interest rate, r, is 4.2%; 4.2%=4.2/100=0.042.
The number of times the interest is compounded annually, n, is 1.
The amount of time, t, is 7.
We know he wants $50,000. This gives us the equation
50000=P(1+0.042/1)</span></span>⁽¹ˣ⁷⁾<span><span>+P{[(1+0.042/1)</span></span>⁽¹ˣ⁷⁾<span><span>]/(0.042/1)}*(1+0.042/1).
Simplifying this a bit, we have
50000=P(1.042)</span></span>⁷<span><span>+P((1.042</span></span>⁷<span><span>)/0.042)*(1.042).
We can factor out P, giving us
50000=P[1.042</span></span>⁷<span><span>+((1.042</span></span>⁷<span><span>)/0.042)*1.042].
This then gives us
50000=P(34.4234).
Divide both sides:
50000/34.4234 = (P(34.4234))/34.4234,
which gives us P=1452.50.</span></span>