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inessss [21]
3 years ago
8

In circle O, the radius is 4, and the length of minor arc AB is 4.2 feet. Find the measure of minor arc AB to the nearest degree

.

Mathematics
2 answers:
PIT_PIT [208]3 years ago
8 0

Answer:

Step-by-step explanation:

Length of a circular arc is given by:

S = rФ

where Ф is the angle in radians subtended at the center by the arc.

Ф = S/r = 4.2 / 4 = 1.05 radians = (1.05*180)/π = 60.16° ≅ 60°

Measure of minor arc AB is 60°

Degger [83]3 years ago
4 0

The formula for arc length is s=r*angle theta where s is the arc length, r is the radius, and angle theta is central angle formed by the arc in radians.

In this case, the angle would be s/r or 4.2/4  which is 1.05 radians. We have to convert this into degrees and so you would multiply 1.05 by (180/pi) which results in approximately 60 degrees. Remember, if you want to convert radians into degrees, the conversion factor is 180/pi and for degrees into radians, it is pi/180.

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A bakery has 456 dozen cookies how many individual cookies are there
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3 years ago
On an alien planet with no atmosphere, acceleration due to gravity is given by g = 12m/s^2. A cannonball is launched from the or
almond37 [142]

Answer:

a) \vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j, b) \theta = \frac{\pi}{4}, c) y_{max} = 84.375\,m, t = 3.75\,s.

Step-by-step explanation:

a) The function in terms of time and the inital angle measured from the horizontal is:

\vec r (t) = [(v_{o}\cdot \cos \theta)\cdot t]\cdot i + \left[(v_{o}\cdot \sin \theta)\cdot t -\frac{1}{2}\cdot g \cdot t^{2} \right]\cdot j

The particular expression for the cannonball is:

\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j

b) The components of the position of the cannonball before hitting the ground is:

x = (90\cdot \cos \theta)\cdot t

0 = 90\cdot \sin \theta - 6\cdot t

After a quick substitution and some algebraic and trigonometric handling, the following expression is found:

0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta}  \right)

0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x

0 = 4050\cdot \sin 2\theta - 6\cdot x

6\cdot x = 4050\cdot \sin 2\theta

x = 675\cdot \sin 2\theta

The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:

\frac{dx}{d\theta} = 1350\cdot \cos 2\theta

1350\cdot \cos 2\theta = 0

\cos 2\theta = 0

2\theta = \frac{\pi}{2}

\theta = \frac{\pi}{4}

Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:

\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta

\frac{d^{2}x}{d\theta^{2}} = -2700

Which demonstrates the existence of the maximum associated with the critical point found before.

c) The equation for the vertical component of position is:

y = 45\cdot t - 6\cdot t^{2}

The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:

\frac{dy}{dt} = 45 - 12\cdot t

45-12\cdot t = 0

t = \frac{45}{12}

t = 3.75\,s

Now, the second derivative is used to check if such solution leads to a maximum:

\frac{d^{2}y}{dt^{2}} = -12

Which demonstrates the assumption.

The maximum height reached by the cannonball is:

y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}

y_{max} = 84.375\,m

7 0
3 years ago
Can some one help with this one question
jolli1 [7]

the answer is D. day 4

5 0
3 years ago
Read 2 more answers
What is the answer to 14x. -2y=46. Y=4x -11
sashaice [31]

Answer:

-42

Step-by-step explanation:

-2y = 46

y = 46/ -2

y = -23

Substitute y in y = 4x-11

-23 = 4x -11

-23+11 =4x

-12 = 4x

x = -12/4

x = -3

Now plug in x value in 14x

14(-3) = -42

5 0
3 years ago
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