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Semenov [28]
3 years ago
15

Pamela is making pizzas. She has 112 pieces of pepperoni. She uses 29 pieces of pepperoni on the first pizza. If she uses the sa

me amount on each pizza, does she have enough pepperoni to make 4 more pizzas?
Mathematics
1 answer:
Svetllana [295]3 years ago
7 0

Answer:

No

Step-by-step explanation:

Number of pieces of pepperoni available with Pamela = 112

Number of pieces of pepperoni used in one pizza = 29

Now, number of pepperoni left after using 29 in one pizza = 112 - 29 = 83

Now, Pamela makes second pizza, so again 29 pieces of pepperoni area used.

Now, number of pepperoni left after using 29 in one pizza = 83 - 29 = 54

Now, Pamela makes third pizza, so again 29 pieces of pepperoni area used.

Now, number of pepperoni left after using 29 in one pizza = 54 - 29 = 25

To make the third pizza, another 29 pepperoni are required but only 25 are available.

Therefore, there is not enough pepperoni for making 4 more pizza.

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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
Which inequality represents the graph below s 75
telo118 [61]

Answer:

Top Left

Step-by-step explanation:

Shaded circle means it will be Greater than or equal to or Less then or equal to. The answwers are less that or equal to 75 so s is less than or equal to 75

5 0
3 years ago
Read 2 more answers
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ira [324]

Hello from MrBillDoesMath!

Answer:

-10w - 20


Discussion:

5 ( -2w - 4 )   =                    

5(-2w) - 5(4) =

-10w - 20

Thank you,

MrB

4 0
3 years ago
Solve the system using substitution x+5y=0 3y+2x=-21
Crazy boy [7]

Answer:

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5 0
2 years ago
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Ann deposits 20% of her earnings each week into her savings account. if she deposited $17 this week, how much did she earn
nika2105 [10]

earnings * 20 percent = 17

earnings * .2 = 17

divide by .2 on each side

earnings = 17/.2

earnings = $85.00

7 0
3 years ago
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