Indra baked and frosted a rectangular cake to sell at a bake sale for the model un club. the cake, including frosting, was 4 inches high, 16 inches wide, and 12 inches long. she centered the cake on a circular tray. the circular tray had a radius of 10 inches. what is the area, to the nearest square inch, of the tray that is not covered by the cake?
a. 100
b. 114
c. 122
d. 314
e. 192

C. 122

Step-by-step explanation:

Given,

Dimension of the cake:

Length = 12 in      Width = 16 in      Height = 4 in

We have to find out the area of the tray that is not covered by the cake.

Indra centered the cake on a circular tray.

Radius of the tray = 10 in.

So area of the tray is equal to π times square of the radius.

Framing in equation form, we get;

Area of tray =

Since the cake is placed in circular tray.

That means only base area of cake has covered the tray.

Base Area of cake =

Area of the tray that is not covered by the cake is calculated by subtracting area of base of cake from area of circular tray.

We can frame it in equation form as;

Area of the tray that is not covered by the cake = Area of tray - Base Area of cake

Area of the tray that is not covered by the cake =

Hence The area of the tray that is not covered by the cake is 122 sq. in.

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

hey seth! : )

i’m not 100% sure, but i think it’s b

hope this ya : p

miss u!

- nicole