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>
True, -4 seems bigger because 4 is bigger than 3 but it's negative and is further away from 0
Answer:
Mean weight = 19 pounds
Step-by-step explanation:
From the question given above, the following data were obtained:
17, 11, 21, 24, 22
Number of data (n) = 5
Mean weight =?
The mean of a set of data is the value obtained by adding all the data together and dividing the result obtained by the total number of data. Thus, the mean can be obtained as follow:
Summation of data = 17+ 11 + 21 + 24 + 22
= 95
Number of data = 5
Mean = Summation of data / Number of data
Mean = 95 / 5
Mean weight = 19 pounds
Therefore, the mean weight of the data is 19 pounds
Answer:
∠BAC = 105°
Step-by-step explanation:
2(5x) = 9x+21
10x = 9x+21
reduce:
x = 21
∠BAC = 5x = 105°
9514 1404 393
Answer:
1 < x < 29
Step-by-step explanation:
The triangle inequality requires the sum of the two shortest sides exceed the longest side.
<u>When x and 14 are the shortest</u>:
x + 14 > 15
x > 1
<u>When 14 and 15 are the shortest</u>:
14 +15 > x
29 > x
Then the requirement for the length of x is ...
1 < x < 29
_____
<em>Additional comment</em>
The length of the third side of a triangle can be between the difference and sum of the two given sides.