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Dima020 [189]
4 years ago
5

22/20 in simplest form

Mathematics
2 answers:
Evgen [1.6K]4 years ago
3 0

Answer: 22/20 in simplest form would be 11/10.

Step-by-step explanation:

To find simplest form, you need to reduce the fraction down by a common factor, for instance 2. 22/2 is 11 (which is your numerator) and 20/2 is 10 (which is your denominator).

Georgia [21]4 years ago
3 0

Answer: 1 and 1/10

Step-by-step explanation: 22÷20= 1.1 = 1 and 1/10

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Given A = {(1, 3)(-1, 5)(6, 4)}, B = {(2, 0)(4, 6)(-4, 5)(0, 0)} and C = {(1, 1)(0, 2)(0, 3)(0, 4)(-3, 5)}, answer the following
Nuetrik [128]

Answer:

  Domain of set B: {2, 4, -4, 0}

Step-by-step explanation:

The domain of the function whose ordered pairs are listed in set B is the set of first numbers of those pairs: {2, 4, -4, 0}.

_____

<em>Comment on the question</em>

A "set" does not have a domain. A "function" has a domain. To make any sense of this question, we have to interpret the question to mean the function described by the ordered pairs in the set.

8 0
3 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
Find the equation of the line that is parallel to the given line and passes through the given point
Sati [7]

Answer:

\huge\boxed{y=-\dfrac{1}{4}x-1\to x+4y=-4}

Step-by-step explanation:

\text{Let}\ k:y=m_1x+b_1;\ l:y=m_2x+b_2\\\\l\ ||\ k\iff m_1=m_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\==========================\\\\\text{We have}\ x+4y=-6.\\\text{Convert to the slope-intercept form:}\\\\x+4y=-6\qquad\text{subtract}\ x\ \text{from both sides}\\\\4y=-x-6\qquad\text{divide both sides by 4}\\\\y=\dfrac{-x}{4}-\dfrac{6}{4}\\\\y=-\dfrac{1}{4}x-\dfrac{3}{2}\to m_1=-\dfrac{1}{4}

\text{Lines are to be parallel. Therefore}\ m_2=-\dfrac{1}{4}.\\\\\text{We initially have the form equation}\ y=-\dfrac{1}{4}x+b.\\\\\text{The line passes through the point}\ (9,\ -3).\\\\\text{Substitute the coordinates of the point to the equation of a line:}\\\\x=9,\ y=-3\\\\-3=-\dfrac{1}{4}(9)+b\\\\-3=-\dfrac{9}{4}+b\qquad\text{add}\ \dfrac{9}{4}\ \text{to both sides}\\\\-\dfrac{12}{4}+\dfrac{9}{4}=b\to b=-\dfrac{3}{4}

\text{Lines are to be parallel. Therefore}\ m_2=-\dfrac{1}{4}.\\\\\text{We initially have the form equation}\ y=-\dfrac{1}{4}x+b.\\\\\text{The line passes through the point}\ (8,\ -3).\\\\\text{Substitute the coordinates of the point to the equation of a line:}

x=8,\ y=-3\\\\-3=-\dfrac{1}{4}(8)+b\\\\-3=-2+b\qquad\text{add 2 to both sides}\\\\-1=b\to b=-1

\text{Therefore the equation is:}\ y=-\dfrac{1}{4}x-1.\\\\\text{Convert to the standard form}\ Ax+By=C:\\\\y=-\dfrac{1}{4}x-1\qquad\text{multiply both sides by 4}\\\\4y=-x-4\qquad\text{add}\ x\ \text{to both sides}\\\\x+4y=-4

5 0
3 years ago
Susan incorrectly factored the expression below. 12a − 15b + 6 3 (4a + 5b + 3)
hichkok12 [17]

Answer:

3(4a−5b+2)

Step-by-step explanation:

The answer should have originally been 3(4a-5b+2)

3(4a+5b+3)= 12a+15b+9 which is incorrect.

6 0
3 years ago
Find k if 3k, k-2, and k + 7 are consecutive terms of an arithmetic sequence.
ahrayia [7]
This is 1:25 (C)

Explanation:

6 0
3 years ago
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