Answer:
(−1.5,1)
Step-by-step explanation:
Finding the distance, midpoint, slope, equation and the x y-intercepts of a line passing between the two points p1 (6,7) and p2 (-9,-5)
The distance (d) between two points (x1,y1) and (x2,y2) is given by the formula
d = √ ((X2-X1)2+(Y2-Y1)2)
d = √ (-9-6)2+(-5-7)2
d = √ ((-15)2+(-12)2)
d = √ (225+144)
d = √ 369
The distance between the points is 19.2093727122985
The midpoint of two points is given by the formula
Midpoint= ((X1+X2)/2,(Y1+Y2)/2)
Find the x value of the midpoint
Xm=(X1+X2)/2
Xm=(6+-9)/2=-1.5
Find the Y value of the midpoint
Ym=(Y1+Y2)/2
Ym=(7+-5)/2=1
The midpoint is: (-1.5,1)
Answer:
(srt(3)/2, -1/2)
Step-by-step explanation:
x = rcos theta
r = 1 on unit circle
x = 1 cos (-pi/6)
x = sqrt(3)/2
y = r sin theta
y =1 sin (-pi/6)=
y = -1/2
At the start, the tank contains A(0) = 50 g of salt.
Salt flows in at a rate of
(1 g/L) * (5 L/min) = 5 g/min
and flows out at a rate of
(A(t)/200 g/L) * (5 L/min) = A(t)/40 g/min
so that the amount of salt in the tank at time t changes according to
A'(t) = 5 - A(t)/40
Solve the ODE for A(t):
A'(t) + A(t)/40 = 5
e^(t/40) A'(t) + e^(t/40)/40 A(t) = 5e^(t/40)
(e^(t/40) A(t))' = 5e^(t/40)
e^(t/40) A(t) = 200e^(t/40) + C
A(t) = 200 + Ce^(-t/40)
Given that A(0) = 50, we find
50 = 200 + C ==> C = -150
so that the amount of salt in the tank at time t is
A(t) = 200 - 150 e^(-t/40)
Substitute in the 3 points to find the values.
<span>P(0,8) = 3(0) + 2(8) = 16 </span>
<span>P(5,4) = 3(5) + 2(4) = 15 + 8 = 23 </span>
<span>P(9,0) = 3(9) + 2(0) = 27 </span>
<span>Therefore (9,0) is the maximum value. </span>
