How much money should be invested now (rounded to the nearest cent), called the initial investment, in a Treasury Bond investment that yields 9% per year, compounded monthly for 10 years, if you wish it to be worth $50,000 after 10 years? Hint: cap a left parenthesis t right parenthesis equals cap a sub o left parenthesis one plus start fraction r over 12 end fraction right parenthesis to the power of 12 t end power comma where A(t) is the final amount, A0 is the initial investment, r is the growth rate expressed as a decimal amount, and t is time in years.
$21,120.54
$5,450.00
$122,567.85
$20,396.8

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!

i honestly dont know

ιт ωιℓℓ вєтнє fιяѕт ¢нσι¢є(α.)

convert the exponential equation to a logarithmic equation using the logarithm base (6) of the right side (36) equals the exponent (2).

ℓσg6(36)=2

нσρє ι ¢συℓ∂ нєℓρ уα

i believe its the last graph