1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
hichkok12 [17]
3 years ago
8

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.

You might be interested in
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)

=5

7 0
2 years ago
Read 2 more answers
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

C te como para el chico del

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
Other questions:
  • Help please!! I’ll give brainliest.....
    7·2 answers
  • Is the following always, sometimes, or never true?
    11·2 answers
  • A. y=82x + 998<br> b. y=247.7x - 335<br> c. y=743x - 1647<br> d. y=998x + 82
    14·2 answers
  • HELP I tried but no luck
    11·1 answer
  • I WILL MAKE BRAINLIEST PLEASE HELP!!!!!!
    14·1 answer
  • 20 Points!!
    14·2 answers
  • (3-x)(x-5)<br> Hurry PLZ
    13·1 answer
  • Giving brainliest!!!!!!!!!!
    5·2 answers
  • Which of the following is the solution to this system of equations?
    10·1 answer
  • Evaluate the quotient of 1.05 x 103 and 1 x 103
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!