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This one may not be nearly as obvious as the previous problem. The given sequence is definitely an arithmetic sequence because there is a common difference. The general formula for an arithmetic sequence is the following:
 

\(a_n=d(n-1)+c\)

d = the common difference

c = the first term in the sequence (sometimes referred to as a₁)

The common difference is the constant amount of change between numbers. In this case, there exists a common differnece.

\(\underbrace{9, 13}\underbrace{, 17},...\\ +4\hspace{3mm}+4\)

Since the common diffence is 4, let's put that in for d.

\(a_n=4(n-1)+c\)

The first term in this sequence is given; it is 9.

\(a_n=4(n-1)+9\)

Just like that, we have generated the formula that generates the nth term in the sequence. To figure out the 48th term, simply evaluate at n=48.

\(a_{48}=4(48-1)+9\\ a_{48}=4*47+9\\ a_{48}=197\)

The nth term is given by:

nth. term = F + (N - 1)*D, where F=First term, N=Number of terms, D=Common difference.

42nd term =7 + (42 - 1)*7

                   =7 + (41*7)

                  =7 + 287

                  =294

The sequence above follows a pattern.

\(f(n)=7n\)

f(n) is the output of the nth term in the sequence

n is the nth term in the sequence.

Since we have established this pattern, we can use this to figure out the 42nd term. If we know that 7n represents the value of the sequence at the nth term in the sequence, then we can use it to find the 42nd term.

\(7*42=294\)

294 is the 42nd term in the sequence. 

Another (similar) answer:

1/2 gal = 2/3 x       Where 'x' = bucket volume       now solve for 'x'

1/2 gal x 3/2 = x = 3/4 gal

I would convert both the multiplicand and multiplier to scientific notation and then multiply.

\(5000\Rightarrow5*10^3\\ 30000\Rightarrow3*10^4\)

Now, multiply them together.

We can solve this with a proportion

(1/2) / (2/3)  =  x  / 1     where x is the gallons  to fill the whole bucket

(1/2)(3/2)  =  x

(3/4)  =  x

𝑓(𝑥) = −2(0.8)^x+1

-2(0.8)^{0+ 1} =  -1.6       not true

-2(0.8)^{-1+ 1}  =  -2 (0.8)⁰  =  -2(1)  =  -2   true

-2(0.8)=^{-1.6+ 1}  =  -2(0.8)^-.6  = -2.286  not true

-2(0.8)^{2.8 + 1}  =  -2 (0.8)^3.8  = -0.856  not true

So......only b is correct

For the second one, we have

11.5x  -  2.3y ≤  20

11.5 (-10)  - 2.3(10) ≤ 20    ⇒ -138  ≤ 20   which is true

11.5 (10)  - 2.3(-10) ≤ 20    ⇒ 138  ≤ 20   which is false

11.5 (0)  - 2.3(5) ≤ 20    ⇒ -11.5  ≤ 20   which is true

11.5 (1)  - 2.3(0) ≤ 20    ⇒ 11.5  ≤ 20   which is true

So....a, c and d are correct

OK.......here's the first one

f(x)  =  -5x  -  8

a. (2,-18)

b. (10, -42)

c. (12,-68)

d. (4,28)

Putting 2 into the funnction we get   -5(2) - 8  =  -18

Putting 10 into the function we have  -5(10) - 8  =  -58

Putting 12 into the function we have  -5(12) - 8  = -68

Putting 4 into the function we have  -5(4) - 8  = -28

So......a and c are correct

Can you tell me how you got the first 3 answers? Thanks

f(5)  =  2(2)^5-4   =  2(2)¹  =  2*2   =  4

The cycle appears to be

Multiply by 3

Add 1

Divide by 2

So.......the series is

3       9     10       5     15     16      8     24      25        12.5

   *3     +1      /2     *3      +1     /2     *3      +1       /2 

For the first one:

 𝑓(𝑥) = −5𝑥 − 8

a. (2,-18)

c. (12,-68)

For the second one :

11.5𝑥 − 2.3𝑦 ≤ 20.8 

a. (-10,10)

c. (0,5)

d. (1,0)

For the third one :

𝑓(𝑥) = −2(0.8)^x+1  

b. (-1,-2)

The explicit rule is  for the nth term  is  :     a_n  =  6*(2)^n-1

Let the number of days  = 365

Then each day  must be  2pi/365  rads

The midline is  [ as before ]    12      and the amplitude here is  15. 8  - 12  =   3.3

I'm going to use the sine function here

I'm letting January 1st   =  "Day 0"...that is......where x  = 0

Using the basic function   y  =  3.3sin [  (2pi* x / 365 ) ]     we need to find the phase shift

Using this function.....one minimum  occurs when x ≈  -91.25 

And we need this minimum to occur on Dec 21st, i.e., 11 units (days) before x = 0 ( Jan 1st) ....so....the phase shift, P,  can be found thusly :

-91.25  + P  =  -11   ⇒    P  =  80.25....so.....this means that we need to shift the graph 80.25 units (days) to the right ........

So......the phase shift  can be expressed as  80.25 *2 *pi  / 365

So....... the [approximate] function  is    y  =  3.3sin [  (2 pi * x / 365)  -  (80.25 * 2pi / 365) ] + 12 

March 27th is the 86th  day of the year..so x  = 85 days after Jan 1st...so the hours of daylight when  x  = 85  are ≈ 12.27  hrs

And October 2nd is the 275th day of the year.....x = 274 days after Jan 1st....so the hours of daylight when   = 274 ≈ 11.365 hrs 

Here is the graph  : https://www.desmos.com/calculator/kyp7tucydf

12,5

cos (60)  =  JL / KL

cos (60)  = JL / 20  ⇒   JL   =  20*cos (60)

And the area can be found as

(1/2) JL * KL * sin (60)   =

(1/2) 20* cos (60) * 20 * sin (60)  =

(1/2) 20 * (1/2) * 20 * sin (60)  =

(1/4)*400 * √3 / 2  =

(1/8)*400*√3  =

50√3    units^2

'A"  is the correct answer for No. 4 

Both of the inequalities  have negative slopes, but the top two graphs have  a line with a positive slope, so the answer cannot be either of these

Looking at the bottom left  graph,  if this one is correct, the point  (0,4)   should make both inequalities true....but notice that this would mean that  4  < (-1/2) (0)  + 2 ⇒  4 < 2     which isn't true

So......the bottom right graph must be correct.....

Draw  perpendicular bisectors to AB, FE and CD....since these are equal chords in the larger circle, by Euclid, equal chords in a circle are equidistant from the center of the larger circle.......

Let O  be this center and let the points where the perpendicular bisectors intersect AB, FE and CD  be labeled as X, Y  and Z, respectively......and....as before, OX  = OY  = OZ  and a smaller circle can  be constructed with  these as radii............but.....this must mean that O  is also the center of this circle, as well....and...  the largest circle that will fit inside the triangle will touch all three sides, so, by definition,  O  is also the incenter of PQR

1260 words per hour   =   1260 words per 60 minutes   =   21 words per minute

Thanks!

A function f has a horizontal asymptote of y = -4, a vertical asymptote of x = 3, and an x-intercept at (1,0). PART A: Let f be of the form

                       f(x)=(ax+b)/(x+c)

                       Find an expression for f(x).

Since we have a vertical asymptote at x = 3, c  = - 3.....and since we have a horizontal asymptote at y  = -4, a  = -4......and since we have an  x intercept at (1,0), we can solve this for b ⇒  -4(1)  +  b =  0 ⇒  b  = 4......here's the graph : https://www.desmos.com/calculator/1pmoaiuapl

PART B: Let f be of the form

                       f(x)=(rx+s)/(2x+t)

Find an expression for f(x).

Since the vertical asymptote is at 3 , we can solve this for t.....2(3) + t  = 0 ⇒  t  = -6

And since the horizontal asymptote is at -4, r  = -8

And since we have an x intercept at (1,0), we can solve this for s......-8(1) + s  = 0 ⇒  s  = 8

Here's a graph : https://www.desmos.com/calculator/gyqo2l6jdl

Thanx , Guest !