Answer:
PΔJKL=66
Step-by-step explanation:
so we are given the line segments JK, KL, and LJ which are tangent to k(O), and also that JA=9, AL=10, and CK=14
JL=JA+AL (parts whole postulate)
JL=9+10=19 (substitution, algebra)
JA=JB=9 (tangent segments from the same point are congruent)
CK=KB=14 (tangent segments from the same point are congruent)
JK=JB+KB (parts whole postulate)
JK=9+14=23 (substitution, algebra)
LA=LC=10 (tangent segments from the same point are congruent)
LK=LC+CK (parts whole postulate)
LK=10+14=24 (substitution, algebra)
Perimeter of ΔJKL=LK+KL+LJ (perimeter formula for triangles)
Perimeter of ΔJKL=23+24+19=66 (substitution, algebra)
Answer:
Step-by-step explanation:
The hypotenuses are equal in length of both triangles.
Answer:
Step-by-step explanation:
3b⁵ + 15b⁴ - 18b⁷ = 3b⁵ + [3*5*b⁴] - [6*3*b⁷]
=3b⁴*( b + 5 - 6b³)
P=2 then 2^2+4=8
P=11 then 11^2+4=125
The value is J = 1.71*10^-6 kmol/m^2.s
According to the given conditions we get,
The length the tube l= 0.20m
The diameter of the tube d= 0.01m
The total pressure inside the tube P= 101.32kPa
The partial pressure of CO2 at the first end is
P1= 456mm Hg = 60794.832 Pa
The partial pressure of CO2 at the other end is
P2= 76mm Hg = 10132.472 Pa
The temperature is T = 298 K
The diffusion coefficient D= 1.67* 10^-5 m^2/s
Generally the molar flux of CO2 is mathematically represented as
J= D{p1-p2} / R.T
Here R is the gas constant with value R= 8.314 j/kmol
So we get J= 1.71 * 10^-3 mol/m^2.sec
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