OM=18, so OQ=QM=18/2=9.
Given QU=8
from figure OQU is a right angled triangle , so OU^2=OQ^2 + QU^2
OU^2 = 9*9 + 8*8 = 81+72=153;
OU=sqrt(153) = 12.37 =13(approx);
From given statements of congruent NT and OU will also be congruent or identical. So, NT=OU=13
So here’s a example and formula
Answer:
x = 50*e∧ -t/100
Step-by-step explanation:
We assume:
1.-That the volume of mixing is always constant 300 gallons
2.-The mixing is instantaneous
Δ(x)t = Amount in - Amount out
Amount = rate * concentration*Δt
Amount in = 3 gallons/ min * 0 = 0
Amount out = 3 gallons/min * x/ 300*Δt
Then
Δ(x)t/Δt = - 3*x/300 Δt⇒0 lim Δ(x)t/Δt = dx/dt
dx/dt = - x/100
dx/ x = - dt/100
A linear first degree differential equation
∫ dx/x = ∫ - dt/100
Ln x = - t/100 + C
initial conditions to determine C
t= 0 x = 50 pounds
Ln (50) = 0/100 * C
C = ln (50)
Then final solution is:
Ln x = - t/100 + Ln(50) or
e∧ Lnx = e ∧ ( -t/100 + Ln(50))
x = e∧ ( -t/100) * e∧Ln(50)
x = e∧ ( -t/100) * 50
x = 50*e∧ -t/100
Answer:
( 3 , -5 )
Step-by-step explanation: