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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

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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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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