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3 years ago

Layla played sports for 3 3/4 hours. Punish time playing sports was 3/8 as long. About how many hours did puno play sports

Mathematics

1 answer:

Nostrana [21]3 years ago

4 0

I would say 13 3/4 but dont take me for my word

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Find all solutions to the following equation. If there are any extraneous solutions, identify them and explain why they are extr

vagabundo [1.1K]

Answer:

b = 3, which is an extraneous solution because if we replace b = 3 in the equation, we must be 5/0 and 30/0, and there is division by 0.

Step-by-step explanation:

$\frac{7}{b+3} + \frac{5}{b-3} = \frac{10b}{b^2-9}$

The minus common multiply must be (b-3)*(b+3) = b² - 9

$\frac{7*(b-3)}{b^2-9} + \frac{5*(b+3)}{b^2 -9} = \frac{10b}{b^2-9}$

7b - 21 + 5b + 15 = 10b

2b = 6

b = 3

Which is an extraneous solution because if we replace b = 3 in the equation, we must be 5/0 and 30/0, and there is division by 0. So this solution is invalid.

7 0

3 years ago

PLEASE HELP GUYS/GIRLS

Lana71 [14]

The answer is B
hope this helps

3 0

3 years ago

S=hp+2b solve for h i forgot the word to decribe how i forgot to do tis

irga5000 [103]

To solve for h, you need to isolate/get h by itself in the equation:

s = hp + 2b Subtract 2b on both sides

s - 2b = hp + 2b - 2b

s - 2b = hp Divide p on both sides to get h by itself

$\frac{s-2b}{p} =\frac{hp}{p}$

$\frac{s-2b}{p} =h$

5 0

3 years ago

Plssss help me, I’ll mark y’all brainliest

My name is Ann [436]

Answer:

1 ray is intersecting point O

Step-by-step explanation:

A ray can be defined as a part of a line that has a fixed starting point but no end point.

AD and EC both go on forever but OB has one endpoint and goes on forever.

3 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

4 years ago