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.
Answer:
1.
Step-by-step explanation:
1. sin 241 = sin(61 + π) = -sin61 = -t^(1/2)
2. sin^2 + cos^2 =1, cos 61= (1-t)^(1/2)
3. [2sin^2 (x)sin(x) +2cos^2 (x)sin(x)]/2cos(x) = 2sin(x) / 2cos(x) = tan(x)
4. tan (x), x do not equal to （π/2 +or- kπ）
x do not equal to 90，270。
Answer:it is only (x,y) -> (3x, 3y)
Step-by-step explanation: I just took the test and got it right,, hope it helped :)
Answer:
The series is absolutely convergent.
Step-by-step explanation:
By ratio test, we find the limit as n approaches infinity of
|[a_(n+1)]/a_n|
a_n = (-1)^(n - 1).(3^n)/(2^n.n^3)
a_(n+1) = (-1)^n.3^(n+1)/(2^(n+1).(n+1)^3)
[a_(n+1)]/a_n = [(-1)^n.3^(n+1)/(2^(n+1).(n+1)^3)] × [(2^n.n^3)/(-1)^(n - 1).(3^n)]
= |-3n³/2(n+1)³|
= 3n³/2(n+1)³
= (3/2)[1/(1 + 1/n)³]
Now, we take the limit of (3/2)[1/(1 + 1/n)³] as n approaches infinity
= (3/2)limit of [1/(1 + 1/n)³] as n approaches infinity
= 3/2 × 1
= 3/2
The series is therefore, absolutely convergent, and the limit is 3/2
![x^{\frac{1}{2}} = \sqrt[2]{x}](https://tex.z-dn.net/?f=%20x%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%3D%20%5Csqrt%5B2%5D%7Bx%7D%20)
and ![3^{-\frac{1}{2}} = \frac{1}{3^{\frac{1}{2}} } = \frac{1}{\sqrt[2]{3}}](https://tex.z-dn.net/?f=%203%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B3%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%5B2%5D%7B3%7D%7D%20)
so
and you rationalize denominator by multiplying numerator and denominatr by 
so that gives--
Your answer is 
