Janelle is planning a party. the cost for 20 people is $290. the cost for 45 people is $590. write the equation in slope intercept form to represent the cost y of having a party for x people.

y = 12x + 50

Step-by-step explanation:

Let y represent the cost of having the party.

Let x represent the number of people who would attend the party.

The equation of a straight line can be represented in the slope-intercept form, y = mx + c

Where c = intercept

Slope, m =change in value of y on the vertical axis / change in value of x on the horizontal axis

change in the value of y = y2 - y1

Change in value of x = x2 -x1

y2 = final value of y

y 1 = initial value of y

x2 = final value of x

x1 = initial value of x

From the information given

y2 = 590

y1 = 290

x2 = 45

x1 = 20

Slope,m = (590 - 290)/(45 - 20) = 300/25 = 12

To determine the intercept, we would substitute x = 20, y = 290 and m= 12 into y = mx + c. It becomes

290 = 12 × 20 + c = 240 + c

c = 290 - 240 = 50

The equation becomes

y = 12x + 50

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)