Answer:
First off, we look for which circles are open or closed.
We start with an open interval since the circle on the left is open and end with a closed interval since the circle on the right is closed.
Domain is all x values, Range is all y values
The graph shows the continous function going from -3 to 1 on the x axis.
According to the circles, this means our domain will be (-3,1].
Now, the range doesn't care about if its closed or not. So we can say the graph is on the y axis from -4 and 0. This means the range is -4<y<0
I used different notations for both just incase you need to represent your answer differently :)
-3<x<1 & (-3,1] . Range is [-4,0]. 0>y>-4 looks correct as-well.
Answer:
a) 1 game
b) 41 goals
c) median = 2
Step-by-step explanation:
a)
As we can see in the line graph, where we have the 0 for the number of goals scored, the graph indicates only 1 in the number of games, so we have only 1 game where no goals were scored.
b)
To find the total number of goals scored, we multiply the goals scored by the number of games for that score, and then sum them all:
total goals = 1*0 + 4*1 + 5*2 + 6*3 + 1*4 + 1*5 = 41 goals
c)
To find the median, we put all the goals in crescent order, and then find the value in the middle. As we have 18 games, the middle value will be an average of the 9th and 10th terms.
We have 1 number 0, 4 numbers 1 and 5 numbers 2 in the beginning, so for these 10 numbers, the 9th and the 10th are the score 2, so the median is 2.
Answer:
Step-by-step explanation:
I gather that you need to find the area of the sector and then subtract from it the area of the segment to get the area of the triangle (although there are other ways in which to find the area of the triangle).
The area of a sector is:
where our angle is given as 34, pi is 3.1415, and the radius is 5:
and if you multiply and divide that all out you get that the area of the sector is:
A = 7.417
Now subtract from it the area of the segment, 2.209. to get the area of the triangle:
7.417 - 2.209 = 5.208
H(t) = -16t² + 60t + 95
g(t) = 20 + 38.7t
h(1) = -16(1²) + 60(1) + 95 = -16 + 60 + 95 = -16 + 155 = 139
h(2) = -16(2²) + 60(2) + 95 = -16(4) + 120 + 95 = -64 + 215 = 151
h(3) = -16(3²) + 60(3) + 95 = -16(9) + 180 + 95 = -144 + 275 = 131
h(4) = -16(4²) + 60(4) + 95 = -16(16) + 240 + 95 = -256 + 335 = 79
g(1) = 20 + 38.7(1) = 20 + 38.7 = 58.7
g(2) = 20 + 38.7(2) = 20 + 77.4 = 97.4
g(3) = 20 + 38.7(3) = 20 + 116.1 = 136.1
g(4) = 20 + 38.7(4) = 20 + 154.8 = 174.8
Between 2 and 3 seconds.
The range of the 1st object is 151 to 131.
The range of the 2nd object is 97.4 to 136.1
h(t) = g(t) ⇒ 131 = 131
It means that the point where the 2 objects are equal is the point where the 1st object is falling down while the 2nd object is still going up.
