<span> D) Train A by a factor of 1.1
</span>Train A: <span>17/35</span><span> = 34.6</span>
<span>Train B: </span>Rate<span> is the </span>slope<span> of the </span>equation, 31.35
Thus,<span>34.6/31.35</span><span> = 1.1036</span>
Answer:
Nutz in my opinion
Step-by-step explanation:
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
Answer:
m<FAB = 75°
m<BAC = 105°
Step-by-step explanation:
First, find the value of x.
(13x - 3)° = (3x + 2)° + 55° (exterior angle theorem of a ∆)
Solve for x
13x - 3 = 3x + 2 + 55
13x - 3 = 3x + 57
Collect like terms
13x - 3x = 57 + 3
10x = 60
Divide both sides by 10
x = 6
✔️m<FAB = 13x - 3
Plug in the value of x
m<FAB = 13(6) - 3 = 78 - 3
m<FAB = 75°
✔️m<BAC = 180 - m<FAB (angles on a straight line/supplementary angles)
m<BAC = 180 - 75 (substitution)
m<BAC = 105°
Answer:
210 mph
Step-by-step explanation:
speed of jet = x
headwind: speed = x - 27
time taken = 1134.6/(x - 27)
tailwind: speed = x + 27
time taken = 1469.4/(x + 27)
same amount of time:
1134.6/(x - 27) = 1469.4/(x + 27)
Cross multiply,
1134.6(x + 27) = 1469.4(x - 27)
1134.6x + 30634.2 = 1469.4x - 39673.8
334.8x = 70308
x = 210 mph
