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3 years ago

Please tell me the correct answer. No link will be acceptable

Mathematics

2 answers:

Nataly [62]3 years ago

8 0

Answer:

The first answer: 5bk + 5b + k

Step-by-step explanation:

Now, you might say it would be the 10bk one, but that's not right.

The only you can get two variables together is through multiplication. That's why 10bk would not work.

inn [45]3 years ago

8 0

The correct answer is 10bk because when you add all the common factors you get that

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Sammie worked at the pizza place last Monday and brought home $60. If he makes $8.50 per hour and earned $13.25 in tips, how man

-BARSIC- [3]

Answer:

5.5 Hours

Step-by-step explanation:

Currently, the equation is 60=8.50h+13.25 and h stands for the hours that Sammie worked. So, you need to solve for h by first subtracting 13.25 (his tips) and getting the amount that he received for the hours that he worked.

You should now have 46.75=8.50h

Our final step, is to divide 46.75 by 8.50 to get h by itself and determine the total hours that Sammie worked.

In the end, you should come out with h=5.5, which means that Sammie worked 5.5 hours.

5 0

2 years ago

Read 2 more answers

in a first aid kit the ratio of large bandages to small bandages is 5 to 2. Based on this ratio, how many large bandages are in

Radda [10]

The best way to answer this question is to set up the first sentence as a ratio of large bandages to total bandages. You would write 5 large bandages to 7 total bandages to 7 total bandages. Then you would make this equivalent to x number of large bandages to 60 total. It would look like this 5:7 = x:60. You could use cross products to multiply 60 by 5 to get 300. 7 times x also should equal 300. Unfortunately, this example will not leave us with a whole number of bandages, but 300/7 is a repeating decimal or a mixed number (42 6/7 large bandages).

5 0

3 years ago

The value of m is equal to three times n divided by the difference of 4p and 5n. which of the equations below is equivalent to t

matrenka [14]

M = (3n) / (4p - 5n) <== ur equation

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

2/3 ÷ 2/12= <br>What is the answer?

Lynna [10]

It’s 4 have a good day

7 0

3 years ago