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Savatey [412]
3 years ago
12

Svp aidez moi jai pas compris cet exercice

Mathematics
1 answer:
Andrej [43]3 years ago
4 0

Answer:

Svp aidez moi jai pas compris cet exercice

Step-by-step explanation:

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What is the answer to -3x + 10 = 5x + -8
Inessa05 [86]

Answer:

x = \frac{9}{4}

Step-by-step explanation:

So here we are trying to find the value of x;

Subtract 5x from both sides;

-8x + 10 = -8

Then, Subtract 10 from both sides;

-8x = -18

Divide both sides by -8;

x = \frac{9}{4}

7 0
3 years ago
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How many more games must Leon play in order to score at least 117 points .
kipiarov [429]

Answer:

how many does he have???

Step-by-step explanation:


8 0
4 years ago
Consider a propeller driven aircraft in forward flight at a speed of 150 mph: The propeller is set up with an advance ratio such
meriva

a)  The thrust of the propeller is  23,834 lbf

b)  The induced velocity is 8.44 mph

c)  The velocity in the downstream far field away from the propeller disc     is 161.56 mph

d) The power absorbed by the fluid is  42.7 hp

e)  the propeller power (thrust flight speed) is  4,844.08 hp

f)  The propeller efficiency is 113.6%

a)The thrust of the propeller can be calculated using the equation,

Thrust = 2πρR^2V^2, where ρ is the air density, R is the propeller radius, and V is the average velocity of the propeller disc. For sea level at a speed of 150 mph, the air density is about 0.002377 slugs/ft^3.

the thrust of the propeller can be calculated as: Thrust = 2π x 0.002377 x (6 ft)^2 x (170 mph)^2 = 23,834 lbf

b)The induced velocity can be calculated using the equation v_ind = (1/2) V ∞ [1–(V_∞/V_tip)^2]^(1/2), where V_∞ is the free stream velocity and V_tip is the tip velocity of the propeller. For the given problem, the induced velocity can be calculated as: v_ind = (1/2) x 150 mph x [1–(150 mph/170 mph)^2]^(1/2) = 8.44 mph

c) The velocity in the downstream far field can be calculated using the equation V_∞ = V_tip – v_ind, where V_tip is the tip velocity of the propeller and v_ind is the induced velocity. For the given problem, the velocity in the downstream far field can be calculated as: V_∞ = 170 mph – 8.44 mph = 161.56 mph

d)The power absorbed by the fluid can be calculated using the equation P_fluid = (1/2) ρ V_ind^3 A_disc, where ρ is the air density, V_ind is the induced velocity, and A_disc is the area of the propeller disc. For the given problem, the power absorbed by the fluid can be calculated as P_fluid = (1/2) x 0.002377 x (8.44 mph)^3 x (π x (6 ft)^2) = 42.7 hp

e)The power absorbed by the propeller can be calculated using the equation P_prop = T x V, where T is the thrust of the propeller and V is the flight speed of the aircraft. For the given problem, the power absorbed by the propeller can be calculated as: P_prop = 23,834 lbf x 150 mph = 3,575,500 ft-lbf/s = 4,844.08 hp

f)The efficiency of the propeller can be calculated using the equation η = P_prop/P_fluid, where P_prop is the power absorbed by the propeller and P_fluid is the power absorbed by the fluid. For the given problem, the efficiency of the propeller can be calculated as: η = 4,844.08 hp/42.7 hp = 113.6%

To know  more  about induced velocity refer to the link brainly.com/question/15280442

#SPJ4

8 0
1 year ago
I only need the first one ( 45% of 30) but please keep in mind that I do need help with the other two, but they are optional.
Usimov [2.4K]
45% of 30 is 13.5
40% of 175 is 70
20% of 60 is 12
6 0
3 years ago
Read 2 more answers
Please help me gow to this bro , im confused
Airida [17]
Ok so remember that a circle is 360 degrees and a line is 180 degrees
so

IKJ
ok so we see that we have a triangle
to find the internal angles (which add to 180)
we do, opposite angles are equal
JIK=59
also CJI+KJI=180 so 132+KJI=180, KJI=48
so
59+48+IKJ=180
73=IKJ




HIB
we can assume that AB is paralell to CD
so therefor IJK=HIB=48 (from previous question)

EIJ is adding to EIH to get 180
EIJ+59=180
EIJ=121

AIK=AIJ+JIK
JIK=59
AIJ=HIB=48
AIK=59+48=107


IKJ=73
HIB=48
EIJ=121
AIK=107
6 0
3 years ago
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