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3 years ago

-3,1,0,1,5 least to greatest

Mathematics

2 answers:

weqwewe [10]3 years ago

8 0

-3015 that is the order

sergey [27]3 years ago

5 0

Least to greatest- -3,0,1,1,5

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Robin puts 58 cups of water into her recipe. Noel puts 13 the amount of water that Robin puts in her recipe.

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So Noel puts 754 cups of water into his/her recipe

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3 years ago

When labeling a graph which comes first the x or y axis

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The first one that comes first is : x axis

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3 years ago

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Packaging By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting fl

Vikki [24]

Answer:

Height: 3/2 inches

Length: 12 inches

Width: 4 inches

Step-by-step explanation:

Let x is the side length of the square

The height of the box by cutting squares off :x

The new length of the cardboard = 15 -2x (because we cut from 4 corners)
The new width of the cardboard = 7 -2x (because we cut from 4 corners)

The new volume of it is:

V = (15 -2x) (7 -2x) x

<=> V = $4x^3 -44x^2 +105x$

To maximum volume, we use the first derivative of the volume

$\dfrac{dV}{dx} = 0$

<=> $12x^2-88x+105=0$

<=> $\left(2x-3\right)\left(6x-35\right) = 0$

<=> 2x -3 = 0 or 6x -35 = 0

<=> x = 3/2 or x = 35/6

To determine which value of x gives a maximum, we evaluate

$\dfrac{d^2V}{dx^2}$ = 24x -88

If x = 3/2, we have:

$\dfrac{d^2V}{dx^2}$ = 24(3/2) -88 = -52

If x = 35/6, we have:

$\dfrac{d^2V}{dx^2}$ = 24(35/6) -88 = 52

We choose x = 3/2 to have the maximum volume because the value of x that gives a negative value is maximum.

So the dimensions (in inches) of the box is:

Height: 3/2 inches

Length: 15-2(3/2) = 12 inches

Width: 7 - 2(3/2) = 4 inches

5 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

Consider the linear function that is represented by the equation y = 2 x + 2 and the linear function that is represented by the

Arisa [49]

Answer:

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers