Which expression is equivalent to
28p^9 q^-5/12p^-6 q^7

7p^15/3q^12 is equivalent to 28p^9 q^-5/12p^-6 q^7

Step-by-step explanation:

Given Parameter:

28p^9 q^-5/12p^-6 q^7

Required; To simplify.

To simplify the above expression, we'll apply 2nd law of indices.

But first, let's rewrite the expression.

28p^9 q^-5/12p^-6 q^7 becomes

(28 * p^9 * q^-5) / (12 * p^-6 * q^7)

Then we collect similar indices. This is done as follows

(28/12) * (p^9/p^-6) * (q^-5/q^7)

From second law of indices (law of division);

If the two terms have the same base () and are to be divided their indices are subtracted.

For instance x^a/x^b = x^(a - b).

Applying this law; we have

(28/12) * (p^9/p^-6) * (q^-5/q^7) becomes

(28/12) * (p^(9 - (-6))) * (q^(-5-7))

(28/12) * (p^(9+6)) * (q^-12)

(28/12) * p^15 * q^-12

Simplify 28/12

(4*7)/(4*3) *p^15 * q^-12

(7/3) * p^15 * q^-12

(7/3) * p^15 * 1/q^12

7p^15/3q^12

Hence, 7p^15/3q^12 is equivalent to 28p^9 q^-5/12p^-6 q^7

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