What is the distance rounded to the nearest tenth between the points(-5, 3) and (1, -1)?
1 answer:
7.2 units is the distance between the two points (-5,3) and (1, -1).
The formula used for determining the distance between two points is shown below. In case this equation is used for finding the distance between the origin and a point, the value of x1 and y1 would just be equal to zero.
distance =
The given points can be directly substituted for the formula.
Let: x2= 1, x1= -5, y2= -1, and y1= 3
distance =
distance =
distance = or 7.21 units
When rounded to the nearest tenth, the distance between the two points would be 7.2 units.
For more information regarding the distance between points, please refer to the link brainly.com/question/23848540.
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see the attached figure with the letters
1) find m(x) in the interval A,B
A (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,C
B(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,B
A (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,C
B(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>
1) find m(x) in the interval A,B
A (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,C
B(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,B
A (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,C
B(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>