I would think that all but one point would be on the line. One way to approach this problem is to find the equation of the line based upon any two points chosen at random, and then determine whether or not the other points satisfy this equation. Next time, would you please enclose the coordinates of each point inside parentheses: (2.5,14), (2.25,12), and so on, to avoid confusion.
14-12
slope of line thru 1st 2 points is m = ---------------- = 2/0.25 = 8
2.50-2.25
What is the eqn of the line: y = mx + b becomes
14 = (8)(2.5) + b; find b:
14-20 = b = -6. Then, y = 8x - 6.
Now determine whether (12,1.25) lies on this line.
Is 1.25 = 8(12) - 6? Is 1.25 = 90? No. So, unless I've made arithmetic mistakes, (1.25, 5) does not lie on the line thru (2.5,14) and (2.25,12).
Why not work this problem out yourself using my approach as a guide?
A)
To be similar triangles have to have equal angles
triangle ZDB'
1)angle Z=90 degrees
triangle B'CQ
1) angle C 90 degrees
angle A'B'Q=90
DB'Z+A'B'Q+CB'Q=180, straight angle
DB'Z+90+CB'Q=180
DB'Z+CB'Q=90
triangle ZDB'
DZB'+DB'Z=180-90=90
DB'Z+CB'Q=90
DZB'+DB'Z=90
DB'Z+CB'Q=DZB'+DB'Z
2)CB'Q=DZB' (these angles from two triangles ZDB' and B'CQ )
3)so,angles DB'Z and B'QC are going to be equal because of sum of three angles in triangles =180 degrees and 2 angles already equal.
so this triangles are similar by tree angles
b)
B'C:B'D=3:4
B'D:DZ=3:2
CQ-?
DC=AB=21
DC=B'C+B'D (3+4= 7 parts)
21/7=3
B'C=3*3=9
B'D=3*4=12
B'D:DZ=3:2
12:DZ=3:2
DZ=12*2/3=8
B'D:DZ=CQ:B'C
3:2=CQ:9
CQ=3*9/2=27/2
c)
BC=BQ+QC=B'Q+QC
BQ' can be found by pythagorean theorem
135
with math any absolute number cant be negative so you just make it positive
Answer:
A:
Misaki had 2 identical pencils that weighed a total of 19 grams, where x is the number of pencils and y is the total weight of the pencils.
Step-by-step explanation:
One of the points is 2 and 19, 2 is along the x axis and 19 is along the y axis