What is the area of a sector with measure of arc equal to 90Á and radius equal to 1 foot? 0.25pi sq. ft. 0.5pi sq. ft. pi sq. ft

2 answers:

5 0

<span>Given a circle with radius r = 8 and a sector with subtended angle measuring 45°, find the area of the sector and the arc length.

They've given me the radius and the central angle, so I can just plug into the formulas. For convenience, I'll first convert "45°" to the corresponding radian value of π/4:

A = ((pi/4)/2)(8^2) = (pi/8)(8^2) = 8pi, s = (pi/4)(8) = 2pi

area A = 8π, arc-length s = 2π
Given a sector with radius r = 3 and a corresponding arc length of 5π, find the area of the sector.

For this exercise, they've given me the radius and arc length. From this, I can work backwards to find the subtended angle.
Then I can plug-n-chug to find the sector area.
5pi = (theta)(3), (theta) = (5/3)pi So the central angle is (5/3)π.
Then the area of the sector is:
A = ((5/3)pi / 2)*(3^2) = ((5/6)pi)*(9) = (15/2)pi A = (15 pi) / 2

</span><span>90/360 = 0.25 pi sq ft</span>

Elis [28]3 years ago

3 0

The area of any sector of a circle can be calculated by multiplying the area of the circle and the angle in degrees over 360. It is expressed as:

Area of a sector = πr²(n/360)

We calculate as follows:

Area of a sector = πr²(n/360)

Area of a sector = π(1²)(90/360)

Area of a sector = 0.25π ft² -----> OPTION 1

You might be interested in

the entire bag is about 800 calories. one serving is 1/5 of the bag. how many calories is in one serving?

Svetradugi [14.3K]

Answer: 160 calories

Step-by-step explanation:

6 0

3 years ago

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows