Ask question

Ask question

All categories

1 year ago

14

Which of the following functions describes the sequence 4, –2, 1, –12, 14, . . .? A. f(1) = 4, f(n + 1) = –2f(n) for n ≥ 1 B. f(

1) = 4, f(n + 1) = f(n) – 2 for n ≥ 1 C. f(1) = 4, f(n) = 12f(n + 1) for n > 1 D. f(1) = 4, f(n) = –12f(n – 1) for n > 1

1 answer:

AlekseyPX1 year ago

5 0

Answer: D. $f(1) = 4,\ \ f(n+1)=\dfrac{-1}{2}f(n)$

Step-by-step explanation:

The given sequence: $4, -2,1,\dfrac{-1}{2},\dfrac14,....$

Here, first term: $f(1)=4$

Second term: $f(2)=-2$

Third term : $f(3)=\dfrac{-1}{2}$

It can be observed that it is neither increasing nor decreasing sequence but having the common ratio.

Common ratio: $r=\dfrac{f(2)}{f(1)}=\dfrac{-2}{4}=\dfrac{-1}{2}$

So, $f(n+1)=\dfrac{-1}{2}f(n)$ [as in G.P. nth term= $a_{n+1}=ar^n$ ]

Hence, correct option is D. $f(1) = 4,\ \ f(n+1)=\dfrac{-1}{2}f(n)$

You might be interested in

On a coordinate plane, a cube root function goes through (negative 2.5, negative 3.5), crosses the y-axis at (0, negative 3), ha

mart [117]

Answer:

$g(x) = \sqrt[3]{x-1} - 2$

Step-by-step explanation:

We want to find h and k in:

$g(x) = \sqrt[3]{x-h} + k$

At the inflection point, the second derivative is equal to zero, so:

$g'(x) = \frac{1}{3} (x-h)^{\frac{-2}{3}}$

$g''(x) = \frac{1}{3} \frac{-2}{3}(x-h)^{\frac{-5}{3}} = 0$

Then x - h = 0.

Inflection point is located at (1, -2), replacing this x value we get:

1 - h = 0

h = 1

We know that the point (-2.5, -3.5) belongs to the function, so:

$-3.5 = \sqrt[3]{-2.5-1} + k$

k ≈ -2

All data, used or not, are shown in the picture attached.

4 0

1 year ago

Find the value of cosAcos2Acos3A...........cos998Acos999A where A=2π/1999

Lady bird [3.3K]

Hello,

Here is the demonstration in the book Person Guide to Mathematic by Khattar Dinesh.

Let's assume
P=cos(a)*cos(2a)*cos(3a)*....*cos(998a)*cos(999a)
Q=sin(a)*sin(2a)*sin(3a)*....*sin(998a)*sin(999a)

As sin x *cos x=sin (2x) /2

P*Q=1/2*sin(2a)*1/2sin(4a)*1/2*sin(6a)*....
   *1/2* sin(2*998a)*1/2*sin(2*999a) (there are 999 factors)
= 1/(2^999) * sin(2a)*sin(4a)*...
   *sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
as sin(x)=-sin(2pi-x) and 2pi=1999a

sin(1000a)=-sin(2pi-1000a)=-sin(1999a-1000a)=-sin(999a)
sin(1002a)=-sin(2pi-1002a)=-sin(1999a-1002a)=-sin(997a)
...
sin(1996a)=-sin(2pi-1996a)=-sin(1999a-1996a)=-sin(3a)
sin(1998a)=-sin(2pi-1998a)=-sin(1999a-1998a)=-sin(a)

So sin(2a)*sin(4a)*...
   *sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
= sin(a)*sin(2a)*sin(3a)*....*sin(998)*sin(999) since there are 500 sign "-".

Thus
P*Q=1/2^999*Q or Q!=0 then
P=1/(2^999)

7 0

1 year ago

Read 2 more answers

answer A radio station located 120 miles due east of Collinsville has a listening radius of 100 miles. A straight road joins Col

melamori03 [73]

Answer:

168.7602 miles

Step-by-step explanation:

One way to solve this problem is by using an equation that describes the listening radius of the station, and another for the road, then the points where this two-equation intersect each other will represent when the driver starts and stops listening to the station, and the distance between the points is the miles that the driver will receive the signal.

The equation for the listening radius (the radio station is at (0,0)):

$x^2+y^2=100^2$

The equation for the road that past through the points (-120,0) and (80,100) (Collinsville and Harmony respectively):

$m=\frac{y_2-y_1}{x_2-x_1} =\frac{100-0}{80-(-120)}=\frac{100}{200}=\frac{1}{2}$

$y-y_1=m(x-x_1)\\y-0=\frac{1}{2}(x-(-120))\\ y=\frac{1}{2}x+60$

Substitutes the value of y in the equation of the circle:

$x^2+(\frac{1}{2}x+60)^2=100^2\\x^2+\frac{1}{4} x^2+60x+3600=10000\\\frac{5}{4} x^2+60x+3600=10000\\\frac{5}{4} x^2+60x+3600-10000=0\\\frac{5}{4} x^2+60x-6400=0\\5 x^2+240x-25600=0\\x^2+48x-5120=0\\$

The formula to solve second-degree equations:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac} }{2a} \\x_{1,2}=\frac{-48\pm\sqrt{48^2-4(1)(-5120)} }{2(1)}\\x_{1,2}=\frac{-48\pm\sqrt{2304+20480} }{2}\\x_{1,2}=\frac{-48\pm\sqrt{22784} }{2}\\x_{1,2}=\frac{-48\pm16\sqrt{89} }{2}\\x_{1,2}=-24\pm8\sqrt{89} \\x_1=-24+8\sqrt{89}\approx51.4718\\x_2=-24-8\sqrt{89}\approx-99.4718\\$

Using the values in x to find the values in y:

$y_1=\frac{1}{2}x_1+60\\y_1=\frac{1}{2}(-24+8\sqrt{89} )+60\\y_1=-12+4\sqrt{89}+60\\ y_1=48+4\sqrt{89}\approx85.7359$

$y_2=\frac{1}{2}x_2+60\\y_2=\frac{1}{2}(-24-8\sqrt{89} )+60\\y_1=-12-4\sqrt{89}+60\\ y_1=48-4\sqrt{89}\approx10.2641$

The distance between the points (51.4718,85.7359) and (-99.4718,10.2641) :

$d=\sqrt{(x_1 -x_2 )^2+(y_1 -y_2)^2} \\d=\sqrt{(-24+8\sqrt{89} -(-24-8\sqrt{89}) )^2+(48+4\sqrt{89} -(48-4\sqrt{89}) )^2}\\d=\sqrt{(-24+8\sqrt{89} +24+8\sqrt{89} )^2+(48+4\sqrt{89} -48+4\sqrt{89} )^2}\\d=\sqrt{(16\sqrt{89} )^2+(8\sqrt{89} )^2}\\d=\sqrt{22784+5696}\\d=\sqrt{28480}\\d=8\sqrt{445}\approx168.7602miles$

4 0

1 year ago

Find pb, if pj=14 and jb= 28

anyanavicka [17]

Answer:

42

Step-by-step explanation:

Based on the information given, point j is the midpoint, as it is shared by both line segments. Set the equation:

pb = pj + jb

Note:

pj = 14

jb = 28

Plug in the corresponding numbers to the corresponding variables:

pb = pj + jb

pb = 14 + 28

pb = 42

pb = 42 is your answer.

~

3 0

1 year ago

Read 2 more answers

A delivery truck is stocked with boxes of leather footballs. The graph shows the relationship between the number of footballs an

pashok25 [27]

Answer:

21

Step-by-step explanation:

To find the answer, you'd have to continue tracing the diagonal line until it intersects with the vertical line that corresponds to the number 3.

If you do that, you'll see that in3 boxes, there are 21 footbals.

You can also calculate it mathematically:

If you have 7 balls per box as shown in the graph, you just have to multiply 7 by 3 to know how many balls you'll find in 3 boxes. 7 * 3 = 21.

Hope it helped,

BioTeacher101

6 0

1 year ago

Read 2 more answers