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patriot [66]
3 years ago
11

Karl starts the engine on his small private airplane. The engine drives a propeller with a radius of 9 feet and its centerline 1

2.5 feet above the ground. At idle, the propeller rotates at a constant speed of approximately 750 revolutions per minute. The height of one propeller tip as a function of time is given by h = 12.5 + 9 sin(750t), where h is the height in feet and t is the time in minutes. Find h when t = 3.5 minutes.
Mathematics
2 answers:
Marta_Voda [28]3 years ago
5 0

Answer:

a=21.1

Step-by-step explanation:

sukhopar [10]3 years ago
4 0

Answer:

  • 3.68 feet using given equation (wrong)
  • 12.5 feet using correct equation

Step-by-step explanation:

You can use the given (incorrect) equation and fill in the value of t to find h:

  h = 12.5 +9sin(750(3.5)) = 3.68 . . . . feet

__

Or, you can use the correct equation, or just your knowledge of revolutions:

  h = 12.5 +9sin(750(2π·3.5)) = 12.5 . . . . feet

in 3.5 minutes at 750 revolutions per minute, the propeller makes 2625 <em>full revolutions</em>, so is back where it started — at 12.5 feet above the ground.

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Find the solution of the system of equations.<br><br>6x+y=30<br><br>−2x−5y=18
IRISSAK [1]

Answer:

Step-by-step explanation:

6x + y = 30

-6x - 15y = 54

-14y = 84

y = -6

6x - 6 = 30

6x = 36

x = 6

(6, -6)

6 0
3 years ago
50 is one tenth of what number?
Stella [2.4K]

Answer:

500

Step-by-step explanation:

Well you find one tenth of a number by dividing it by 10

So to do the opposite, you can do 50*10 which is 500

Or if you really wanted to you could go through each option dividing them all by 10 till you got 50

5 0
3 years ago
Read 2 more answers
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Naddik [55]
I think the answer is $1.56





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8 0
2 years ago
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = e^−2
Mekhanik [1.2K]

Answer:

x=1

y=s

z=1

Step-by-step explanation:

(x, y, z)=(1, 0, 1)

Substitute 0 for y

y = e^{-2t} *sin 4t\\0 = e^{-2t} *sin 4t\\\\0 = e^{-2t} *(\frac{e^{4t}-e^{-4t}}{2j} )\\0 = e^{-2t} *(e^{4t}-e^{-4t} )\\0 = e^{-2t} *e^{4t}*(1-e^{-8t} )\\\\0 = e^{2t}*(1-e^{-8t} )\\\\\\Either   \\0 = e^{2t}\\t = -inf\\or\\0 = (1-e^{-8t} )\\(e^{-8t} ) = 1\\ -8t = ln(1) =0\\t=0\\\\

Confirming if t=0 satisfy the other equation

x = e^−2t cos 4t = e^−2(0)cos(4*0)

= e^(0)cos(0) = 1

z = e^−2t  = e^−2(0)  = 0

Therefore t=0 satisfies the other equation

Finding the tangent vector at t=0

\frac{dx}{dt}=-2te^{-2t} cos4t + e^{-2t}(-4sin4t)=-2(0)e^{-2(0)} cos4(0) + e^{-2(0)}(-4sin4(0))=0\\\\ \frac{dy}{dt} =-2te^{-2t} sin4t + e^{-2t}(4cos4t)=-2(0)e^{-2(0)} sin4(0)+ e^{-2(0)(4cos4(0)) }= 1\\\\\frac{dz}{dt}  = -2te^{-2t}  = -2(0)e^{-2(0)}  = 0

The vector equation of the tangent line is

(1, 0, 1) +s(0,1,0)= (1, s, 1)

The parametric equations are:

x=1

y=s

z=1

6 0
3 years ago
How do I solve for x and y?
Lesechka [4]
Cancel out the terms from one side first then do it for the other one. 4 example make x and individual by multiply the other side by 9 the evaluate. do the same for y 
7 0
3 years ago
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