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AlexFokin [52]
3 years ago
10

What is 3x3x8x help it’s so hard pls

Mathematics
1 answer:
Vesna [10]3 years ago
8 0
72 if your needing to times them
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What is a real life word problem for (-7) + (7). Solve problem. Can you help?
notka56 [123]

Answer:

Step-by-step explanation:

Sally owes 7 dollars to Matthew. She opens her piggy bank and takes out 7 dollars. Now, how much money does she owe to Matthew?

-7 + 7 = 0

-7 + 7 is the same as  7 - 7

You can clearly see that 7-7 is 0.

3 0
3 years ago
Read 2 more answers
ANSWER ASAP PLEASEEEEEE
expeople1 [14]

Answer:

the answer is A

Step-by-step explanation:

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8 0
2 years ago
Read 2 more answers
Figure out what the percent is
DaniilM [7]

The percentage of data that is roughly greater than 66, as displayed in the box plot, is 100%.

<h3>How to Determine a Percentage of a Data Represented in a Box Plot?</h3>

In a box plot, we have the following displayed five-number summary which tells what percentage of the data distribution for each part of the data distribution:

Upper quartile (Q3): This is the value at where the box in the box plot ends at the edge of the box. From this point to the left, all data values that fall within the bracket make up 75% of the data.

Lower quartile (Q3): This is the value at where the box in the box plot starts at the edge of the box. From this point to the left, all data values that fall within the bracket make up 25% of the data.

Median: this is the middle value at the point where the line divides the box and data below this point make up 50% of the data.

The other five-number summary are the maximum and the minimum values that are represented by the whiskers.

On the box plot given, 66 is at the extreme whisker at the left. This means that the percentage of data that is roughly greater than 66 is 100%.

Learn more about the box plot on:

brainly.com/question/14277132

#SPJ1

7 0
2 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
4 years ago
Will make brainiest if answered correctly<br><br><br> write the value of the expression<br> 2^3/2^3
Lesechka [4]
1 because it’s the answer
8 0
4 years ago
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