OLYMPICS The number of men and women participating in the Winter

A rational function can have infinitely many x-values at which it is not continuous. True or False. If false, explain

sattari [20]

<span>It is false since the rational function is discontinuous when the denominator is zero. But the denominator is a polynomial and a polynomial has only finitely many zeros. So the discontinuity points of a rational function is finite. </span>

3 0

1 year ago

Margo has 15 pounds of clay, Tim has 10 pounds of clay, Tray has 13 pounds of clay, Lila has 14 pounds of clay, and Jerry has 13

weqwewe [10]

Add all the clay together and divide it by 5 so the answer is 13

4 0

2 years ago

Read 2 more answers

Which statement correctly describes the slope of the linear function that is represented by the data in the table? x y 8 –8 8 –4

spin [16.1K]

Answer:

Step-by-step explanation:

slope=change in y/change in x

one way is t pick 2 points, (x1,y1) and (x2,y2)

the slope is (y2-y1)/(x2-x1)

pick some points

(8,4) and (8,0)

slope=(0-4)/(8-8)=-4/0=undefined

the slope is undefined

there is no slope.

5 0

1 year ago

Read 2 more answers

Show that the Fibonacci numbers satisfy the recurrence relation fn = 5fn−4 + 3fn−5 for n = 5, 6, 7, . . . , together with the in

Sonja [21]

Answer with step-by-step explanation:

We are given that the recurrence relation

$f_n=5f_{n-4}+3f_{n-5}$

for n=5,6,7,..

Initial condition

$f_0=0,f_1=1,f_2=1,f_3=2,f_4=3$

We have to show that Fibonacci numbers satisfies the recurrence relation.

The recurrence relation of Fibonacci numbers

$f_n=f_{n-1}+f_{n-2}$ , $f_0=0,f_1=1$

Apply this

$f_n=(f_{n-2}+f_{n-3})+f_{n-2}=2f_{n-2}+f_{n-3}$

$f_n=2(f_{n-3}+f_{n-4})+f_{n-3}=3f_{n-3}+2f_{n-4}$

$f_n=3(f_{n-4}+f_{n-5})+2f_{n-4}=5f_{n-4}+3f_{n-5}$

Substitute n=2

$f_2=f_1+f_0=1+0=1$

$f_3=f_2+f_1=1+1=2$

$f_4=f_3+f_2=2+1=3$

Hence, the Fibonacci numbers satisfied the given recurrence relation .

Now, we have to show that $f_{5n}$ is divisible by 5 for n=1,2,3,..

Now replace n by 5n

$f_{5n}=5f_{5n-4}+3f_{5n-5}$

Apply induction

Substitute n=1

$f_5=5f_1+3f_0=5+0=5$

It is true for n=1

Suppose it is true for n=k

$f_{5k}=5f_{5k-4}+3f_{5k-5}$ is divisible 5

Let $f_{5k}=5q$

Now, we shall prove that for n=k+1 is true

$f_{5k+5}=5f_{5k+5-4}+3f_{5k+5-5}=5f_{5k+1}+3f_{5k}=5f_{5k+1}+3(5q)$

$f_{5k+5}=5(f_{5k+1}+3q)$

It is multiple of 5 .Therefore, it is divisible by 5.

It is true for n=k+1

Hence, the $f_{5n}$ is divisible by 5 for n=1,2,3,..

8 0

2 years ago

Use the position function s(t) = −16t² + 400, which gives the height (in feet) of an object that has fallen for t seconds from a

skad [1K]

Answer:

160m/s

Step-by-step explanation:

The object can hit the ground when t = a; meaning that s(a) = s(t) = 0

So, 0 = -16a² + 400

16a² = 400

a² = 25

a = √25

a = 5 (positive 5 only because that's the only physical solution)

The instantaneous velocity is

v(a) = lim(t->a) [s(t) - s(a)]/[t-a)

Where s(t) = -16t² + 400

and s(a) = -16a² + 400

v(a) = Lim(t->a) [-16t² + 400 + 16a² - 400]/(t-a)

v(a) = Lim(t->a) (-16t² + 16a²)/(t-a)

v(a) = lim (t->a) -16(t² - a²)(t-a)

v(a) = -16lim t->a (t²-a²)(t-a)

v(a) = -16lim t->a (t-a)(t+a)/(t-a)

v(a) = -16lim t->a (t+a)

But a = t

So, we have

v(a) = -16lim t->a 2a

v(a) = -32lim t->a (a)

v(a) = -32 * 5

v(a) = -160

Velocity = 160m/s

7 0

1 year ago