Answer:
The sum of the internal ángles = 360°
(3y+40)° and (3x-70°) are suplementary angles = 180°
then:
(3x-70) + (3y+40) + 120 + x = 360 ⇒ first eq.
(3y+40) + (3x-70) = 180 ⇒ second eq
development:
from the first eq.
3x + x + 3y = 360 + 70 - 40 - 120
4x + 3y = 430 - 160
4x + 3y = 270 ⇒ third eq.
3y = 270 - 4x
y = (270 - 4x) / 3 ⇒ fourth eq.
from the secon eq.:
3y + 3x = 180 + 70 - 40
3y + 3x = 250 - 40
3y + 3x = 210 ⇒ fifth eq.
multiply by -1 the fifth eq and sum with the third eq.
-3y - 3x = -210 ⇒ (fifth eq. *-1)
3y + 4x = 270
⇒ 0 + x = 60
x = 60°
from the fourth eq.
y = (270-4x)/3
y = (270-(4*60)) / 3
y = (270 - 240) / 3
y = 30/3
y = 10°
Probe:
from the first eq.
(3x-70) + (3y+40) + 120 + x = 360
3*60 - 70 + 3*10 + 40 + 120 + 60 = 360
180 - 70 + 30 + 40 + 120 + 60 = 360
180 + 30 + 40 + 120 + 60 - 70 = 360
430 - 70 = 360
Answer:
y = 10
The answer is A. 6x^2 hope this helps!
The minimum distance will be along a perpendicular line to the river that passes through the point (7,5)
4x+3y=12
3y=-4x+12
y=-4x/3+12/3
So a line perpendicular to the bank will be:
y=3x/4+b, and we need it to pass through (7,5) so
5=3(7)/4+b
5=21/4+b
20/4-21/4=b
-1/4=b so the perpendicular line is:
y=3x/4-1/4
So now we want to know the point where this perpendicular line meets with the river bank. When it does y=y so we can say:
(3x-1)/4=(-4x+12)/3 cross multiply
3(3x-1)=4(-4x+12)
9x-3=-16x+48
25x=51
x=51/25
x=2.04
y=(3x-1)/4
y=(3*2.04-1)/4
y=1.28
So now that we know the point on the river that is closest to Avery we can calculate his distance from that point...
d^2=(x2-x1)^2+(y2-y1)^2
d^2=(7-2.04)^2+(5-1.28)^2
d^2=38.44
d=√38.44
d=6.2 units
Since he can run at 10 uph...
t=d/v
t=6.2/10
t=0.62 hours (37 min 12 sec)
So it will take him 0.62 hours or 37 minutes and 12 seconds for him to reach the river.
It looks like your question is incomplete. I believe you also have options to pick which graph is correct. However, I can still give you the information you are looking for.
The slope of the line is -1.2
The Y-intersect is (0, 4)
I have also attached an image of what the graph would look like.
Hope this helps.
