Answer:
See Below
Step-by-step explanation:
1) 180 ft / 6 sec
To find a unit rate, you need to have a denominator of 1 and to to do that, you can divide the numerator and denominator by the same number, in this case: 6.
30 ft / 1 sec
2) 86 mi / 5 hr
Same as above^(except dividing by 5)
17.2 mi / 1 hr
3) 83 mi / 5 hr
Same as above^^
16.6 mi / 1 hr
We can rewrite this as: 16.6 mph
4) 93 mi / 5 hr
Same as above^^^
18.6 mi / 1 hr
5) 150 mi / 4 hr
Same as above^^^^
37.5 mi / 1 hr
We can rewrite this as: 37.5 mph
Answer:
5
Step-by-step explanation:
The slope is 5 because the equation is y = 200-5x where 5 is the slope because it is the rate of change.
We are given a graph of a quadratic function y = f(x) .
We need to find the solution set of the given graph of a quadratic function .
<em>Note: Solution of a function the values of x-coordinates, where graph cut the x-axis.</em>
For the shown graph, we can see that parabola in the graph doesn't cut the x-axis at any point.
It cuts only y-axis.
Because solution of a graph is only the values of x-coordinates, where graph cut the x-axis. Therefore, there would not by any solution of the quadratic function y = f(x).
<h3>So, the correct option is 2nd option :∅.</h3>
see the attached figure with the letters
1) find m(x) in the interval A,BA (0,100) B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100
2) find m(x) in the interval B,CB(50,40) C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20
3)
find n(x) in the interval A,BA (0,0) B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x
4) find n(x) in the interval B,CB(50,60) C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30
5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find <span>h'(x)
</span>h'(x)=-36/25=-1.44
6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x)
h'(x)=18/25=0.72
for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72
<span> h'(x) = 1.44 ------------ > not exist</span>