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3 years ago

Y=3x+2 i need a graph for this

Mathematics

1 answer:

klasskru [66]3 years ago

7 0

Graph is in the picture.

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Find the perimeter of the colored part of the figure. The figure is composed of small squares with side-length 1 unit and curves

harina [27]

Answer:

16.56 in

Step-by-step explanation:

4 sides of the squares are exposed

The radius of each semicircle is 1 since they are against 2 squares.

2pi*r=circumfrence

2pi*1=6.28

6.28/2=3.14 because it's a semicircle

3.14*4=12.56 because there are 4 equal semicircles

12.56+4=16.56

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

3 years ago

How do i write a number in standerd form

pickupchik [31]

In its normal way, For example: 214, Write it in the way it is Usually written.

8 0

3 years ago

Read 2 more answers

Find the value of k so that 48x-ky=11 and (k+2)x+16y=-19 are perpendicular lines.

Rufina [12.5K]

Answer: k = -1 +/- √769

Step-by-step explanation:

48x - ky = 11

-48x -48x

-ky = -48x + 11

$\frac{-ky}{-k} = \frac{-48x}{-k} + \frac{11}{-k}$

$y =\frac{48x}{k} - \frac{11}{k}$

Slope: $\frac{48}{k}$

*************************************************************************

(k + 2)x + 16y = -19

- (k + 2)x -(k + 2)x

16y = -(k + 2)x - 19

$\frac{16y}{16} = -\frac{(k + 2)x}{16} - \frac{19}{16}$

$y = -\frac{(k + 2)x}{16} - \frac{19}{16}$

Slope: $-\frac{(k + 2)}{16}$

**********************************************************************************

$\frac{48}{k}$ and $-\frac{(k + 2)}{16}$ are perpendicular so they have opposite signs and are reciprocals of each other. When multiplied by its reciprocal, their product equals -1.

$-\frac{(k + 2)}{16}$ * $\frac{k}{48}$ = -1

$\frac{(k + 2)k}{16(48)}$ = 1

Cross multiply, then solve for the variable.

(k + 2)(k) = 16(48)

k² + 2k - 768 = 0

Use quadratic formula to solve:

k = -1 +/- √769

5 0

3 years ago

Graph of f(x) = 100(0.7)x

Lana71 [14]

Answer:

pay attention in class :)

Step-by-step explanation:

7 0

3 years ago