All categories

3 years ago

What is the arc length if Θ = 2pi/3 and the radius is 4 cm? 8pi/3 16pi/3 pi/6 3pi/4

2 answers:

5 0

There's a simple formula for finding the arc length.

In radians:
A = θ*r = (2pi/3) x 4 cm = 8*pi/3 cm

Therefore, the answer is letter A 8*pi/3 cm.

I hope my answer has come to your help. God bless and have a nice day ahead!

Triss [41]3 years ago

3 0

The arc length would be 8pi/3

You might be interested in

What procedures could you use to show that the sun's path in the sky changes over a year? Choose the three correct answers. A. M

Verizon [17]

Answer:

Its A and D

Step-by-step explanation:

I found that out the hard way

6 0

3 years ago

What does 19+r= PLEASE HELP

NemiM [27]

Okay so since there aren’t any values that r is equal to 19 plus r is equal to 19r. I’m pretty sure that’s the answer

4 0

3 years ago

Find the interval(s) over which the function is increasing

anastassius [24]

The intervals refer to the x values where the function (graph) is increasing which means the y values are getting larger from left to right. Parantheses are used not brackets (parantheses means it doesn't include the value and brackets mean it does) so here the intervals would be
( negative infinity , -3)U( -3, 0 )

3 0

3 years ago

What’s is the answer to this equation -3 - 5n = 1 - 6n

Tomtit [17]

-3 - 5n = 1 - 6n

+6n +6n

-3 + n = 1

+3 +3

n = 4

Hope this helps! ;)

4 0

3 years ago

Given f'(x) = (x-4)(4-2x), find the x-coordinate for relative maximum on the graph of f(x)

antiseptic1488 [7]

Answer:

The coordinate of the relative maximum is x=4.

Step-by-step explanation:

Given that the derivative of the function $f(x)$ is $f'(x) = (x-4)(4-2x)$ , the maxima and minima or the critical points can be found where $f'(x)=0,$ that is:

$f'(x) = (x-4)(4-2x)=0.$

The solutions to this equation are $x=4$ and $x=2.$

Now, if the second derivative $f''(x)$ for a function $f(x)$ is negative at a critical point, then the critical point is the relative maximum.

Therefore we want to see at which of the two critical points is $f''(x)$ negative. The second derivative is:

$\frac{\mathrm{d f'(x)}}{\mathrm{d}x}=f''(x)=-4(x-3).$

Now $f''(4)$ is $-4$ , and $f''(2)$ is $+4$ , therefore we deduce that the relative maxium is located at $x=4$ , because there the second derivative $f''(x)$ is negative.

4 0

3 years ago