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andrew-mc [135]
3 years ago
6

The two most commonly used systems of coordinates are the rectangular (or Cartesian) and the

Mathematics
2 answers:
IgorC [24]3 years ago
7 0
The answer is B quadrant 

Vikentia [17]3 years ago
7 0

Answer:

D. Polar.

Step-by-step explanation:

We are asked to choose the two most commonly used systems of coordinates.

We know that most commonly used systems of coordinates are the rectangular coordinate system and the polar coordinate system.

The rectangular coordinate system is also known as Cartesian coordinate system.

Therefore, option D is the correct choice.

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The yearbook club is handing out T-shirts to its members. There are 5 blue, 7 green, 9 red, and 4 yellow T-shirts in all. If Jac
Neporo4naja [7]

Answer:

well it's 9/25 or 0.36% or (36%) only

Step-by-step explanation:

So you add all the numbers up it would be 25, divide the numerator by the denominator and get 0.36%, i can't remember if you divide .36 by 100 or not but anyway you get 36%.

But I hope i have helped you in anyway.

5 0
4 years ago
What is the slope of the equation y = 2 x - 1? 3 x​
N76 [4]

Answer:

2

Step-by-step explanation:

If the equation is in the form y = mx + b, then m is the slope and b is the y-intercept (For example - in the equation y = 3x + 2 the slope is 3 and the y-intercept is 2)

In this case, the number next to x is 2, so that's the slope.

4 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
What three-dimensional solid has 6 rectangular faces, 2 equal bases that are not rectangles, and 18 edges?
natima [27]
A hexagonal prism should work! The two bases are hexagons, and there are 6 lateral faces that are rectangles. When you count the edges, there are 18... here's an image of one (just in case)

3 0
3 years ago
Read 2 more answers
Help me answer this plz I really need help
expeople1 [14]

Answer:

Answer E. is correct

Step-by-step explanation:

b = 3 * c --> c = 1/3 * b

c + 3 = 1/3 * b + 3 --> E.

4 0
4 years ago
Read 2 more answers
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