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Airida [17]
1 year ago
8

Express the radical using the imaginary unit, iii. Express your answer in simplified form.

Mathematics
1 answer:
Reika [66]1 year ago
4 0

iii\implies i^3\implies i^2\cdot i\implies (-1)\cdot \sqrt{-1}\implies -\sqrt{-1}\implies -i

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christa went shopping for back to school clothes.Before the tax was added on, her total was $176. if the sales tax is 6% what wa
skelet666 [1.2K]
<span>Christa went shopping for back to school clothes.
She paid a total of 176 dollars before the tax was added on.
Now, how much will she be paying if the sales tax is 6% what was the total amount she paid.
Solutions:
=> 176 dollars * 6% to get the amount
=> 6% / 100% = 0.06
=> 176 dollars * .06 = 10.56
Now, let’s add to get the answer
=> 176 dollars + 10.56 dollars = 186.56 dollars</span>



7 0
4 years ago
Read 2 more answers
For the function defined by f(t)=2-t, 0≤t&lt;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
8 coins to equal $1.06
SVEN [57.7K]

Answer: 3 quarters is equal to .75 and 2 dimes is equal to .20

So far we have .95 with 5 coins

2 nickels which is equal to .10

1 penny

8 coins = 1.06

Step-by-step explanation:

7 0
3 years ago
The coefficient of determination is used to: compute correlations for truncated ranges. compare the relative strength of coeffic
kozerog [31]

Answer: compare the relative strength of coefficients.

Step-by-step explanation: The Coefficient of determination usually denoted as R^2 is obtained by taking the squared value of the correlation Coefficient (R). It's value ranges from 0 to 1 and the value obtained gives the proportion of variation in the dependent variable which could be attributed to it's correlation or relationship to th independent variable. With a R^2 value close to 1, this means a large portion of Variation in a variable A could be explained due to changes in variable B while a low value signifies a low variance between the variables. Hence, the Coefficient of determination is used in comparing the relative strength of the Coefficients in other to establish whether a weak or strong relationship exist.

6 0
3 years ago
Round off to the nearest <br> 3,142
Bezzdna [24]
Just put 3.150 lol that what I did
5 0
3 years ago
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