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4 years ago

(b) Solve for d. Show each step for full credit. Circle your final answer.

Mathematics

1 answer:

andrew11 [14]4 years ago

3 0

The answer is d = g/(e-f) which you can write as $d = \frac{g}{e-f}$

note: don't forget the parenthesis around the "e-f" terms if you choose to write it the first way

check out the attached image for the steps done

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Jacques deposited $1900 into an account that earns 4% interest. After t year, jaques has $3875.79in the account. How long was th

Sladkaya [172]

Deposit = $1900
Amount After interest = $3875.79
% interest = 4%
Interest = 1975.79

Time For Interest Period

4/100 × 1900 × t = 1975.79
76 t = 1975.79
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3 years ago

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madreJ [45]

The expression as given doesn't make much sense. I think you're trying to describe an infinitely nested radical. We can express this recursively by

$\begin{cases}a_1=\sqrt{42}\\a_n=\sqrt{42+a_{n-1}}\end{cases}$

Then you want to know the value of

$\displaystyle\lim_{n\to\infty}a_n$

if it exists.

To show the limit exists and that $a_n$ converges to some limit, we can try showing that the sequence is bounded and monotonic.

Boundedness: It's true that $a_1=\sqrt{42}\le\sqrt{49}=7$ . Suppose $a_k\le 7$ . Then $a_{k+1}=\sqrt{42+a_k}\le\sqrt{42+7}=7$ . So by induction, $a_n$ is bounded above by 7 for all $n$ .

Monontonicity: We have $a_1=\sqrt{42}$ and $a_2=\sqrt{42+\sqrt{42}}$ . It should be quite clear that $a_2>a_1$ . Suppose $a_k>a_{k-1}$ . Then $a_{k+1}=\sqrt{42+a_k}>\sqrt{42+a_{k-1}}=a_k$ . So by induction, $a_n$ is monotonically increasing.

Then because $a_n$ is bounded above and strictly increasing, the limit exists. Call it $L$ . Now,

$\displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n-1}=L$

$\displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty}\sqrt{42+a_{n-1}}=\sqrt{42+\lim_{n\to\infty}a_{n-1}}$

$\implies L=\sqrt{42+L}$

Solve for $L$ :

$L^2=42+L\implies L^2-L-42=(L-7)(L+6)=0\implies L=7$

We omit $L=-6$ because our analysis above showed that $L$ must be positive.

So the value of the infinitely nested radical is 7.

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