Answer:
=3
Step-by-step explanation:
following the order of operations, whatever is in the parentheses gets done first. so you would add 13 + 2 and get 15. then you would subtract 9-4 and get 5. then you would divide the numbers that you got and get 3.
-6 And 8 ok it make me have 20 c
Answer:
(5, 1) and (-5, -7)
Step-by-step explanation:
Given the simultaneous equations
x^2– y^2 = 24 .... 1
x= 3 + 2y .... 2
Substitute 2 into 1
(3-2y)^2-y^2 = 24
Expand
9 - 12y + 4y^2 - y^2 = 24
9 - 12y + 3y^2 = 24
3y^2 - 12y + 9-24 = 0
3y^2 -12y -15 = 0
y^2-4y-5 = 0
y^2+5y - y - 5 = 0
y(y+5) - 1 (y+5) = 0
y-1 = 0 and y+5 = 0
y = 1 and -5
Substitute the values of y into 2
Recall that x = 3+3y
when y = 1
x = 3 + 2(1)
x = 5
when x = -5
x = 3 + 2(-5)
x = 3 - 10
x = -7
Hence the solutions are (5, 1) and (-5, -7)
Answer:
P_he = 562.5 mmHg
P_ne = 1068.75 mmHg
P_ar = 618.75 mmHg
Step-by-step explanation:
Formula for partial pressure is;
Partial pressure = mole fraction × total pressure
Mole fraction of Helium(He); x_he= 1/(1 + 1.9 + 1.1) = 1/4 = 0.25
Mole fraction of Neon(Ne); x_ne = 1.9/(1 + 1.9 + 1.1) = 1.9/4 = 0.475
Mole fraction of argon; x_ar = 1.1/(1 + 1.9 + 1.1) = 1.1/4 = 0.275
Thus;
Partial pressure of He;
P_he = 0.25 × 2250
P_he = 562.5 mmHg
Partial pressure of Ne;
P_ne = 0.475 × 2250
P_ne = 1068.75 mmHg
Partial pressure of Ar;
P_ar = 0.275 × 2250
P_ar = 618.75 mmHg
Answer:
440 hours
Step-by-step explanation:
In this case the aircraft has completed one cycle of service in the first 454 hours of service. Then making the difference between the hours of service and the hours of the first cycle we have:
Hours of the second cycle= Total time in service- time Airworthiness Directive
=468-454=14
Then we have 14 hours of service and need to accumulate the rest to complete 454. It is
454-14=440
Then we need to accumulate 440 hours additional to complete the Airworthiness Directive .
