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Mrac [35]
3 years ago
9

What is the maximum height the bee reaches as it flies from A to B?

Mathematics
1 answer:
Natasha2012 [34]3 years ago
5 0
It's asking you to find the derivative, which is -0.06x+1.8, and then, when it equals zero, it has reached the top. that is, -0.06x+1.8=0, or x= 30. plug that into the original formula, and you get -0.03(30)²+1.8(30)-24, or 3. the answer is 3 (inches, presumably)
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The Farrells save $150 each month for their next summer vacation.Write an equation that they can use to find ,s, their savings a
Andrei [34K]

Answer:

<u>s=150m</u>

Step-by-step explanation:

Using slope intercept form, you can come p with the equation, <u>s=150m</u> where 150 is the savings per month and you were not given a starting point which would be y.

y=mx+b

<u>s=150m</u>

7 0
3 years ago
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What is the radius of the circle given by the equation below? x2 + 14x + y2 - 14y = -73
PolarNik [594]

Answer:

5

Step-by-step explanation:   5/13  8 48

graph the function, center the center of the circle find the difference between the center and the circle

center at (-7, 7)

sorry I don't know how to find it mathematically.  I'll try to figure out later, if you need an mathematically solution.  

8 0
3 years ago
Write the prime factorization of 42. Use exponents when appropriate and order the factors from least to greatest
jeka94
42: 2,3,7
2 x 3 x 7 = 42

4 0
3 years ago
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Forces of 9 lbs and 13 lbs act at a 38º angle to each other. Find the magnitude of the resultant force and the angle that the re
fiasKO [112]

Answer: R=20.84\ lb\quad 22.57^{\circ},15.43^{\circ}

Step-by-step explanation:

Given

Two forces of 9 and 13 lbs acts 38^{\circ} angle to each other

The resultant of the two forces is given by

\Rightarrow R=\sqrt{a^2+b^2+2ab\cos \theta}

Insert the values

\Rightarrow R=\sqrt{9^2+13^2+2(9)(13)\cos 38^{\circ}}\\\Rightarrow R=\sqrt{81+169+184.394}\\\Rightarrow R=\sqrt{434.394}\\\Rightarrow R=20.84\ lb

Resultant makes an angle of

\Rightarrow \alpha=\tan^{-1}\left( \dfrac{b\sin \theta}{a+b\cos \theta}\right)\\\\\text{Considering 9 lb force along the x-axis}\\\\\Rightarrow \alpha =\tan^{-1}\left( \dfrac{13\sin 38^{\circ}}{9+13\cos 38^{\circ}}\right)\\\\\Rightarrow \alpha =\tan^{-1}(\dfrac{8}{19.244})\\\\\Rightarrow \alpha=22.57^{\circ}

So, the resultant makes an angle of 22.57^{\circ} with 9 lb force

Angle made with 13 lb force is 38^{\circ}-22.57^{\circ}=15.43^{\circ}

7 0
3 years ago
Read 2 more answers
A 4-lb. force acting in the direction of (vector) 4,-2 moves an object just over 7ft from point (0,4) to (5,-1). Find the work d
Tcecarenko [31]

To solve this problem, we have to find the net displacement and the net force and the multiply the dot product together and get the work done.

The work done on moving the object is 27ft*lbs

<h3>Work done in moving the object from point A to point B</h3>

To find the work done on this object, let's find the net force on the object.

Data;

  • force = 4lb
  • direction = 4, -2
  • displacement = 7ft
  • direction = (0, 4) to (5,1)

The unit vector of the force is

\sqrt{4^2 +(-2)^2} =\sqrt{16 + 4} = \sqrt{20}

\frac{4}{\sqrt{20} }, \frac{-2}{\sqrt{20} }

The net force acting on the object is

F = 4(\frac{4}{\sqrt{20} }, \frac{-2}{\sqrt{20} })\\F= (\frac{16}{\sqrt{20} }, \frac{-8}{\sqrt{20} } )

The displacement on the object is 7ft through (0,4) to (5, -1)

The unit vector on displacement is

\sqrt{5^2 + (-1-4)^2} = \sqrt{25+25} = \sqrt{50}

\frac{5}{\sqrt{50} }, \frac{-5}{\sqrt{50} }

The net displacement will be

7(\frac{5}{\sqrt{50} }, \frac{-5}{\sqrt{50} }) = \frac{35}{\sqrt{50} }, \frac{-35}{50}

The work done will be F.d

w = f. d \\

w = (\frac{16}{\sqrt{20} }, \frac{-8}{\sqrt{20} } ) * \frac{35}{\sqrt{50} }, \frac{-35}{50}\\w = 17.71+ 8.854\\w = 26.567 = 27ft*lbs

The work done on moving the object is 27ft*lbs

Learn more on work done on an object here;

brainly.com/question/26152883

#SPJ1

6 0
3 years ago
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