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andre [41]
3 years ago
11

Is this correct?I just started learning this​

Mathematics
1 answer:
MrRa [10]3 years ago
6 0
2^3 = 8
3^2 = 9
5^2 = 25
2^5 = 32

(5^2 = 5*5 = 25)
(2^5 = 2*2*2*2*2 = 32)
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Find the missing side length​
grin007 [14]

Answer:

11

Step-by-step explanation:

it's 10 because on the image it gives you the 7 which would be the top half of the side and the 4 would be the bottom so you just and then and get 11

6 0
4 years ago
15 POINTS IF YOU CAN ANSWER THIS CORRECTLY!!!
My name is Ann [436]

4√2x−8+4√2x+8=0

Simplifies to:

11.313708x=0

Let's solve your equation step-by-step.

11.313708x=0

Step 1: Divide both sides by 11.313708.

<u>11.313708x </u>

11.313708

=

<u>       0        </u>

11.313708

x=0

Answer:

x=0

4 0
4 years ago
How do i find the y-intercept of an eqaution that passes throught two points
Pachacha [2.7K]

They way to find the y-intercept from two points is to first find the slope, which can be derived from the formula m=\frac{y_2-y_1}{x_2-x_1}.

Then, you can plug in one of the two points (x_1,y_1) or (x_2,y_2) into the formula y=mx+b to solve for b, which is your y-intercept.


3 0
4 years ago
Why is <img src="https://tex.z-dn.net/?f=lim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7Bx%7D%7Bln%20x%7D" id="TexFormula1" title="li
aleksklad [387]

The first limit is infinity since x>\ln(x) for all x>0.

The second limit is zero since 2^x converges to 0 as x\to-\infty, while x^2 is very large and positive.

8 0
2 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
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