4(5a²b³)^2 / (2x³y^5)4 show your work

you invest $15,000 in a savings account with an annual interest rate of 2.5% in which the interest is compounded quarterly. how much money should you expect to have in the account after 5 years? show your work

Step-by-step explanation:

1) 4(5a²b³)^2 / (2x³y^5)4

Opening the parenthesis in the denominator, it becomes

4(5a²b³)^2 / 8x³y^5

Recall: (b^x)^y = b ^(xy)

It becomes

4(5^2a^4b^6) / 8x³y^5

4×25a^4b^6) / 8x³y^5

= 100a^4b^6) / 8x³y^5

2) Initial amount invested into the account is $15000 This means that the principal is P, so

P = 15000

It was compounded quarterly. This means that it was compounded 4 times in a year. So

n = 4

The rate at which the principal was compounded is 2.5%. So

r = 2.5/100 = 0.025

It was compounded for 5 years. So

t = 5

The formula for compound interest is

A = P(1+r/n)^nt

A = total amount in the account at the end of t years. Therefore

A = 15000 (1+0.025/4)^4×5

A = 15000 (1+0.00625)^20

A = 15000 (1.00625)^20

A = 16990.62

Approximately $16991

a) $ 9.50

b) 17

c) y-intercept (0, 12). this is the initial amount of money that joe had prior to playing any game.

d) x-intercept ( 24, 0). this is the number of games that joe will play before he depletes the initial amount of money that he had.

e) slope is -0.5. joe's balance will reduce by 0.5 for every game that he plays.

f)

domain; [0,24]

range; [0,12]

g) x + 2y = 24

step-by-step explanation:

a)

if joe plays 5 games, this will imply that x = 5. substitute x = 5 in the equation and solve for y;

y = -.50x + 12

y = -0.50(5) + 12

y = 9.50

b)

if joe leaves the arcade with $3.50, this will mean that y = 3.50. plug in y = 3.50 in the given equation and solve for x;

y = -.50x + 12

3.50 = -0.50x + 12

0.50x = 12 - 3.50

0.5x = 8.5

x = 17

c)

the function given is   y = -.50x + 12.

the y-intercept is the point where the graph of this function will intersect the y-axis. at this point, the value of x is usually 0. plug in x = 0 in the given equation and determine y;

y = -0.50(0) + 12

y = 12

the y-intercept is thus (0, 12). this is the initial amount of money that joe had prior to playing any game.

d)

the x-intercept refers to the point where the graph of a function intersects with the x-axis. at this point, the value of y is 0. substitute y = 0 in the given equation and solve for x;

0 =   -.50x + 12

0.5x = 12

x = 24

x-intercept ( 24, 0). this is the number of games that joe will play before he depletes the initial amount of money that he had.

e)

the slope of the given equation is -0.5. the equation has been given in slope-intercept form. thus the slope is simply the coefficient of x.

joe's balance will reduce by 0.5 for every game that he plays. this is the interpretation of the slope.

f)

domain is the set of x values for which the function is real and defined. the domain will thus be; [0,24]. this will restrict the balance to only assume positive values.

on the other hand, the range is the set of y values for which a function is real and defined. the range in our case will be [0,12]. joe can have between $0 and $12 balance.

g)

the standard form of the equation of a straight line is;

ax + by = c

where a, b, and c are integers.

our equation is given as;

y = -.50x + 12

re-writing the equation we have;

0.5x + y = 12

divide all through by 0.5;

x + 2y = 24