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vladimir1956 [14]
3 years ago
12

Please help me out with this!!!!! Help is much needed !

Mathematics
1 answer:
Anit [1.1K]3 years ago
5 0

here the centre of circle(h,k) is (-2,4)

and its radius(r) is 6 uints.

now the equation of circle is,

(x-h)^2 +(y-k)^2 =r^2

or, (x+2)^2 + (y-4)^2=6^2

or, x^2 + 4x + 4+y^2 -8y+16=36

or, x^2 +y^2 + 4x -8y -16=o

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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.)

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4 years ago
What would be the simple interest on an amount of $4,000 at 4% interest?
iris [78.8K]

Answer:

the answer would be $160

7 0
3 years ago
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Please help with #1 and explain your answers! Thank you
sashaice [31]

1a) False. A square is never a trapezoid. A trapezoid has only one pair of parallel sides while the other set of opposite sides are not parallel. Contrast this with a square which has 2 pairs of parallel opposite sides.

1b) False. A rhombus is only a rectangle when the figure is also a square. A square is essentially a rhombus and a rectangle at the same time. If you had a Venn Diagram, then the circle region "rectangle" and the circle region "rhombus" overlap to form the region for "square". If the statement said "sometimes" instead of "always", then the statement would be true.

1c) False. Any rhombus is a parallelogram. This can be proven by dividing up the rhombus into triangles, and then proving the triangles to be congruent (using SSS), then you use CPCTC to show that the alternate interior angles are congruent. Finally, this would lead to the pairs of opposite sides being parallel through the converse of the alternate interior angle theorem. Changing the "never" to "always" will make the original statement to be true. Keep in mind that not all parallelograms are a rhombus.

8 0
3 years ago
Read 2 more answers
real estate julie bought a house for $100,00 five years ago. if the value of the house has appreciated 5% per year , how much is
Stels [109]
To find the answer, turn the appreciated percent into a decimal:

5% ----> .05 (Divide by 100)

Then multiply it by the total value:

100,000 x .05 = 5000

Then add that to the total:

105,000

So the house is worth a total of $105,000.

Hope this helps!
5 0
4 years ago
The graph of the equation y = į x+2 is displayed.
Oksi-84 [34.3K]

Answer:

first

third and

fourth

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