Enter this matrix in matlab: >> f = [0 1; 1 1] use matlab to find an invertible matrix p and a diagonal matrix d such that pdp-1 = f. use matlab to compare f10 and pd10p-1. let f = (1, 1)t. compute ff, f2f, f3f, f4f, and f5f. (you don't need to include the input and output for these.) describe the pattern in your answers. consider the fibonacci sequence {fn} = {1, 1, 2, 3, 5, 8, 13…}, where each term is formed by taking the sum of the previous two. if we start with a vector f = (f0, f1)t, then ff = (f1, f0 + f1)t = (f1, f2)t, and in general fnf = (fn, fn+1)t. here, we're setting both f0 and f1 equal to 1. given this, compute f30.

1. 8

2. 9

3. 20

4. 3

5. 17

6. 2.2

7. 11.25

8. 1

9. 14

10. 2

step-by-step explanation:

thats the first half

a.   we have two

lines:   y = 4-x   and  

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b.   4-x = 8-x^-1

  x           4-x       8-x^-1

-3           7        

8.33

-2           6         8.5

-1           5         9

  0           4        

-

  1           3        

7

  2           2        

7.5

  3           1         7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:  

x=-4.236

y = 4-x   = 4 – (-4.236)

= 8.236

y = 8-x^-1 =   .236)^-1

= 8.236

x=0.236

  y = 4-x   = 4 – (0.236) = 3.764

y = 8-x^-1 =   8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c.   to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x   and   y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines.   the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!

x = 3 ± √(5+y)