2(n+3)= 2n+3 explain whys it weong ?

The graph of f(x)=3•2^x -3 is shown below. G(x) is a transformation of f(x). How would you write the equation for the function g

Bezzdna [24]

Answer:

C

Step-by-step explanation:

To find the appropriate equation, notice the red graph has shifted up about 6 units. This means that $3*2^x-3$ which is shifted by -3 will move 6 up. -3+6= 3. This means the function will have +3. Only B and C are options. Now test a point. Notice the red function crosses the y-axis at about 5 or 6 when x=0.

$g(x) = 2^x+3\\g(0) = 2^0+3\\g(0)=1+3\\g(0)=4$

This function doesn't match.

$f(0) = 3*2^0+3\\f(0) = 3*1+3\\f(0)=6$

This function matches. Option C is correct.

3 0

3 years ago

How to do question 5 and 6. Can you explain the first one which is A for 5 and 6? Thank you

Illusion [34]

for 6

in a

given

slope(m)=1

point=(2,-4)=(x1,y1)

we know

equation of straight line

y-y1=m(x-x1)

y+4=1(x-2)

y+4=x-2

therefore x-y-6=0 is the required equation

substituting with ax+by+c=0 we get

a=1

b= -1

c= -6

4 0

3 years ago

Can u guys plz give me the answer to number 1-3

jek_recluse [69]

I'm pretty sure it is 5:55

8 0

3 years ago

If g(x)= x+1/x-2 and h(x)= 4-x, what is the value of (g•h)(-3)?

jek_recluse [69]

Answer:

8/5

Step-by-step explanation:

First find h(-3)

h(-3)= 4-(-3) = 4+3 = 7

Then take this result and find g(h(-3)) = g(7)

g(7) = (7+1)/( 7-2) = 8/5

8 0

3 years ago

Given a joint PDF, f subscript X Y end subscript (x comma y )equals c x y comma space 0 less than y less than x less than 4, (1)

ioda

(1) Looks like the joint density is

$f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0$

In order for this to be a proper density function, integrating it over its support should evaluate to 1. The support is a triangle with vertices at (0, 0), (4, 0), and (4, 4) (see attached shaded region), so the integral is

$\displaystyle\int_0^4\int_y^4 cxy\,\mathrm dx\,\mathrm dy=\int_0^4\frac{cy}2(4^2-y^2)=32c=1$

$\implies\boxed{c=\dfrac1{32}}$

(2) The region in which X > 2 and Y < 1 corresponds to a 2x1 rectangle (see second attached shaded region), so the desired probability is

$P(X>2,Y$

(3) Are you supposed to find the marginal density of X, or the conditional density of X given Y?

In the first case, you simply integrate the joint density with respect to y:

$f_X(x)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dy=\int_0^x\frac{xy}{32}\,\mathrm dy=\begin{cases}\frac{x^3}{64}&\text{for }0$

In the second case, we instead first find the marginal density of Y:

$f_Y(y)=\displaystyle\int_y^4\frac{xy}{32}\,\mathrm dx=\begin{cases}\frac{16y-y^3}{64}&\text{for }0$

Then use the marginal density to compute the conditional density of X given Y:

$f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}\frac{2xy}{16y-y^3}&\text{for }y$

6 0

3 years ago