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3 years ago

Please assist me with these problems.

Mathematics

1 answer:

timurjin [86]3 years ago

6 0

Answer:

c for the question that says what point is on $y=\log_a(x)$ given the options.

9 for the question that reads: "If $\log_a(9)=4$ , what is the value of $a^4$ .

Step-by-step explanation:

We are given $y=\log_a(x)$ .

There are some domain restrictions:

$a \text {is number between } 0 \text{ and } 1 \text{ or greater than } 1$

$x \ge 0$

a) couldn't be it because x=0 in the ordered pair.

b) isn't is either for the same reason.

c) \log_a(1)=0 \text{ because } a^0=1[/tex]

So c is so far it! Since (x,y)=(1,0) gives us $0=\log_a(1)$ where the equivalent exponential form is as I mentioned it two lines ago.

d) Let's plug in the point and see: (x,y)=(a,0) implies $0=\log_a(a)$ .

The equivalent exponetial form is $a^0=a$ which is not true because $a^0=1 (\neq a)$ .

If $\log_a(9)=4$ . then it's equivalent exponential form is: $a^4=9$ .

Guess what it asked for the value of $a^4$ and we already found that by writing your equation $\log_a(9)=4$ in exponential form.

Note:

The equivalent exponential form of $\log_a(x)=y$ implies $a^y=x$ .

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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

You make a $15 payment on your loan of $500 at the end of each month

lara [203]

Answer:

2.8 years or 33.6 months.

Step-by-step explanation:

I am not sure what your questions is, but I assume it is how long it will take to pay it off?

In a year (15*12,) you would have paid $180 of it.

x = 500/180

Therefore, it will take you approximately 2.8 years to pay off your loan, excluding interest, of course, since you did not provide that rate.

8 0

2 years ago

2.5y=5 what is the answer to the equation?

postnew [5]

Answer:2

Step-by-step explanation:

2.5(2)

7 0

2 years ago

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100 POINTS!!!!!!

Andrews [41]

Answer:

A 4x2=7-15 X=

Step-by-step explanation:

B como dedos de los dedos de losdos para el desayuno

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4 0

2 years ago

The length of a rectangle is 10 yards more than the width. If the perimeter is 68 yards, what are the length and width?

quester [9]

Divide the perimeter by 2, this will give you the sum of one length and one width.

68/2 = 34 yards

Now the length is 10 yards more, so subtract 10 from the 34 and divide by 2 again:

34 - 10 = 24

24 /2 = 12

The width is 12 yards.

The length is 12 + 10 = 22 yards.

5 0

3 years ago

Please assist me with these problems.​

Please assist me with these problems.