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3 years ago

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Newton currently has a balance of $1,716.18 in an account he has held for 29 years. He opened the account with an initial deposi

t of $784. What is the simple interest rate on the account?

1 answer:

elixir [45]3 years ago

3 0

The simple interest rate is 4.1%.

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At the yearly fund raising drive, the nonprofit company 's goal was to raise $55,500 each day. After three days it had raised $5

maxonik [38]

This is honestly simple but we all need help! The answer is 55,053. Hope this helps! =3

5 0

3 years ago

For the function defined by f(t)=2-t, 0≤t<1, sketch 3 periods and find:

Oksi-84 [34.3K]

The half-range sine series is the expansion for $f(t)$ with the assumption that $f(t)$ is considered to be an odd function over its full range, $-1$ . So for (a), you're essentially finding the full range expansion of the function

$f(t)=\begin{cases}2-t&\text{for }0\le t$

with period 2 so that $f(t)=f(t+2n)$ for $|t|$ and integers $n$ .

Now, since $f(t)$ is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

$f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L$

where

$b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt$

In this case, $L=1$ , so

$b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt$
$b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{4-2(-1)^n}{n\pi}$

The half-range sine series expansion for $f(t)$ is then

$f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t$

which can be further simplified by considering the even/odd cases of $n$ , but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming $f(t)$ is even/symmetric across its full range. In other words, you are finding the full range series expansion for

$f(t)=\begin{cases}2-t&\text{for }0\le t$

Now the sine series expansion vanishes, leaving you with

$f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L$

where

$a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt$

for $n\ge0$ . Again, $L=1$ . You should find that

$a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3$

$a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt$
$a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}$
$a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}$

Here, splitting into even/odd cases actually reduces this further. Notice that when $n$ is even, the expression above simplifies to

$a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0$

while for odd $n$ , you have

$a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}$

So the half-range cosine series expansion would be

$f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t$
$f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t$
$f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t$

Attached are plots of the first few terms of each series overlaid onto plots of $f(t)$ . In the half-range sine series (right), I use $n=10$ terms, and in the half-range cosine series (left), I use $k=2$ or $n=2(2)-1=3$ terms. (It's a bit more difficult to distinguish $f(t)$ from the latter because the cosine series converges so much faster.)

5 0

4 years ago

Need help answering this math problem <br> X-5/4=-9/5 then x =

Mrac [35]

I’m pretty sure the answer is -1

5 0

3 years ago

Read 2 more answers

I would like to know the value of b and c

Ratling [72]

Answer:

a = 3, b = 8, c = 14

Step-by-step explanation:

First, we can organize this, putting lower values first.

2, 3, 3, 5, 9, 9, 11, 12

a , b , c somewhere there

Given this information, one thing we can look at first is the mode. The mode is 3, so there must be more 3s than any other number. Currently, there are 2 3s and 2 9s, so there must be at least one more 3 and no more 9s to make that true. Therefore, a, b, or c is 3. Therefore, we have

2, 3, 3, 3, 5, 9, 9, 11, 12

2 of a, b, c somewhere in there

Next, the median is 8. In our current state, the median is 5. There are 9 numbers, with 11 total including the 2 remaining values. Because there will be an odd amount of values, the median must be a number on the list. Therefore, our list is

2, 3, 3, 3, 5, 8, 9, 9, 11, 12

1 of a, b, c somewhere in there

There are 4 numbers above the median and 5 numbers below right now. To balance this out, there must be another number above the median. As a consequence, the remaining value must be greater than 8.

Finally, we know that the range is 12, so maximum - minimum = 12. Because the remaining number must be greater than 8, the minimum number is 2, no matter what. Therefore,

maximum - 2 = 12

add 2 to both sides to isolate maximum

maximum = 14

There is no 14 currently on the list, so the remaining value must be 14.

Our a, b, and c are as follows, in order from smallest to largest:

3, 8, 14

8 0

2 years ago

What is the difference between the solution of a linear

Snowcat [4.5K]

A system of linear inequalities in two variables consists of at least two linear inequalities in the same variables. The solution of a linear inequality is the ordered pair that is a solution to all inequalities in the system and the graph of the linear inequality is the graph of all solutions of the system.

4 0

3 years ago