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hram777 [196]
3 years ago
5

What is the solution to the equation? 3/4(4c+16)=2c+9

Mathematics
2 answers:
guajiro [1.7K]3 years ago
7 0
\frac{3}{4}(4c+16)=2c+9\ \ \ |use\ distributive\ property\ a(b+c)=ab+ac\\\\\frac{3}{4}\cdot4c+\frac{3}{4}\cdot16=2c+9\\\\\frac{3}{\not4_1}\cdot\frac{\not4c}{1}+\frac{3}{\not4_1}\cdot\frac{\not16^4}{1}=2c+9\\\\3c+12=2c+9\ \ \ \ |subtract\ 2c\ and\ 12\ from\ both\ sides\\\\\boxed{c=-3}
ella [17]3 years ago
4 0
<span>3/4(4c+16)=2c+9

We simplify the equation to the form, which is simple to understand
<span>3/4(4c+16)=2c+9

Reorder the terms in parentheses
<span>+(+3c+12)=2c+9

Remove unnecessary parentheses
<span>+3c+12=+2c+9

We move all terms containing c to the left and all other terms to the right.
<span>+3c-2c=+9-12

We simplify left and right side of the equation.
<span>+1c=-3

We divide both sides of the equation by 1 to get c.
<span>c=-3
Hope this helped! :)

</span></span></span></span></span></span></span>
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For the function defined by f(t)=2-t, 0≤t&lt;1, sketch 3 periods and find:
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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

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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.)

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lisabon 2012 [21]

Answer:

That is good! Just plug them in the graph and you should be good. Also this should be a parabola correct?

Step-by-step explanation:

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