1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
nekit [7.7K]
3 years ago
10

Which statement about the inequality 215<235 is true? On the vertical number line, 215 is located below 235. On the vertical

number line, 235 is located below 215. On the vertical number line, 235 and 215 are located at the same point. On the vertical number line, 215 and 235 are both located below 0.
Mathematics
1 answer:
masha68 [24]3 years ago
4 0
On vertical line 215 is located below 235
You might be interested in
Find the slope between the two given points: (-2,-1) and (-4,-7)<br><br><br> pls helppp
Zielflug [23.3K]
-7+1/-4+2 is -6/-2 which is 3
6 0
2 years ago
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
3 years ago
Write a function in any form that would match the graph shown below.
disa [49]

Answer:

0, 5

Step-by-step explanation:

I am not shure that is right but I think an ordered pair is form of a funtion.

6 0
3 years ago
Solve the following inequality -6 &gt; - 3c
Inessa05 [86]
Dividing both sides by-3
the division sign will still be the same
your answer is c>2
4 0
3 years ago
The average of Shondra's test scores in physics is between 88 and 93 what is the inequality
kodGreya [7K]

Answer:

The inequality is

88<x<93

Step-by-step explanation:

The average of Shondra's test scores in physics is between 88 and 93.

Let me give out the meaning of some inequality symbols

<= Less than or equal to

>= Greater than or equal to

< Less than

> Greater than

Let the average score be x

In this case , the average score is between 88 and 93

The inequality is

88<x<93

3 0
3 years ago
Other questions:
  • Find the x and y intercept of the line 4x+2y=24
    6·1 answer
  • Write the sum as a product. simplify the product (-2)+(-2)+(-2)+(-2)
    7·2 answers
  • How many two digit natural numbers division by 7​
    8·1 answer
  • Find the zeroes of 6x^4+x^3+2x^2-4x+1
    12·1 answer
  • The terminal ray of a 300° angle lies in the<br> This angle measures<br> i radians.<br> quadrant.
    15·2 answers
  • What is equivalent to 8/12
    14·1 answer
  • HELP ME IF RIGHT ILL MARK YOU BRAINLIEST
    7·1 answer
  • Deltamath <br><br> Please help me
    6·2 answers
  • Which property explains why these two expressions are equal?
    12·1 answer
  • How many 10 cm pieces of wood would you need to make one big piece that is 1 meter<br> long?
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!