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Sever21 [200]
3 years ago
11

How do you evaluate a function?

Mathematics
1 answer:
GrogVix [38]3 years ago
8 0
Substitute the input (the given number or expression) for the function's variable (place holder, x). Replace the x with the number or expression. Given thefunction f (x) = 3x - 5, find f (4). Solution: Substitute 4 into the functionin place of x.
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<img src="https://tex.z-dn.net/?f=%5Csqrt%7B272%7D" id="TexFormula1" title="\sqrt{272}" alt="\sqrt{272}" align="absmiddle" class
lozanna [386]

√272 is a real number, and the simplified expression of √272 is 4√17

<h3>How to simplify the expression?</h3>

The expression is given as:

√272

Expand

√272 = √16 * 17

Take the square root of 16

√272 = 4√17

Hence, the simplified expression of √272 is 4√17

Read more about real numbers at:

brainly.com/question/7784687

#SPJ1

7 0
1 year ago
How to find the surface area
Readme [11.4K]
Are you talking about area ?
3 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
Which describes a non-uniform probability model? A) heads or tails from a coin flip B) the sum result of rolling two dice C) sel
irga5000 [103]

Answer:

Option B.

Step-by-step explanation:

The non-uniform probability is when the event is not same in the occurring of event.

Option A is uniform because there will be equal chance of getting   head or tail.

Option B is non-uniform because sum can be any number  if we get(3,4) it will be 7 and if we have (2,3) then it will be 5.

Option C is uniform because we have equal number of red and green balls so, selecting a ball from both will be same.

Option D is uniform because chance of 1,2,,3,4,5 or 6 will be equal chance from rolling a die that is 1/6 in all cases.

Therefore, Option B is correct

7 0
3 years ago
Read 2 more answers
Which graph shows the solution to the system of linear inequalities?
Brilliant_brown [7]

Answer:

this is answer

Step-by-step explanation:

To graph a linear inequality in two variables (say, x and y ), first get y alone on one side. Then consider the related equation obtained by changing the inequality sign to an equality sign. The graph of this equation is a line. If the inequality is strict ( < or > ), graph a dashed line.

4 0
2 years ago
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