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.
1.208 ounce because 30.2 divided by 25 equals 1.208 ounces
It’s A because I did the test lmafo
Answer:
C
If x = o then f(0) = 4(2) * 0 = 0
If f(x) = 4(2) 0 - 13
then f(x) = -13 at x = 0
Answer:
0.64
Step-by-step explanation:
P(J / R) = P (J and R) / P(R)
0.8 = P (J and R) / 0.6
P (J and R) = 0.6 * 0.8 = 0.48 [Probability John practicing and it is raining]
P(J / NR) = P (J and NR) / P(NR)
0.4 = P (J and NR) / (1 - 0.6) = P (J and NR) / 0.4
P (J and NR) = 0.4 * 0.4 = 0.16 [Probability John practicing and it is not raining]
Hence;
Probability of John practicing regardless of weather condition is
P(John Practicing) = 0.48 + 0.16 = 0.64
HOPE THIS HELPED!!!
