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3 years ago

8

If the perimiter of a sector is 4 times its radius, then the radian measure of the ceenral angel of the sector is

1 answer:

BigorU [14]3 years ago

6 0

The perimeter of a sector includes two segments that are each equal to the radius, and the arc length. For your sector, the arc length is found as
perimeter = 4r = r + r + arc-length
2r = arc-length

Now we also know that arc-length is related to the central angle (in radians) by
arc-length = r*central-angle
2r = r*central-angle
2 = central-angle

The measure of the central angle of the sector is 2 radians.

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Answer:

D.(1, -1, 2) hope this helps

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3 years ago

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55. (a) If alpha and beta are the roots of the equation xsquare+ px+q=0 and beta>alpha find the square of the

densk [106]

Answer:

√(p²-4q)

Step-by-step explanation:

Using the Quadratic Formula, we can say that

x = ( -p ± √(p²-4(1)(q))) / 2(1) with the 1 representing the coefficient of x². Simplifying, we get

x = ( -p ± √(p²-4q)) / 2

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x = ( -p + √(p²-4q)) / 2 and x = ( -p - √(p²-4q)) / 2. The difference of the roots is thus

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3 years ago

What is the y - intercept of the equation y = 8x+4?<br> x-1

Tomtit [17]

Answer:

4

Step-by-step explanation:

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4 years ago

cylinder shaped can needs to be constructed to hold 200 cubic centimeters of soup. The material for the sides of the can costs 0

Dafna11 [192]

Answer:

Radius=2.09 cm

Height,h=14.57 cm

Step-by-step explanation:

We are given that

Volume of cylinderical shaped can=200 cubic cm.

Cost of sides of can=0.02 cents per square cm

Cost of top and bottom of the can =0.07 cents per square cm

Curved surface area of cylinder= $2\pi rh$

Area of circular base=Area of circular top= $\pi r^2$

Total cost,C(r)= $0.02\times 2\pi rh+2\pi r^2\times 0.07$

Volume of cylinder, $V=\pi r^2 h$

$200=\pi r^2 h$

$h=\frac{200}{\pi r^2}$

Substitute the value of h

$C(r)=0.02\times 2\pi r\times \frac{200}{\pi r^2}+2\pi r^2\times 0.07$

$C(r)=\frac{8}{r}+0.14\pi r^2$

Differentiate w.r.t r

$C'(r)=-\frac{8}{r^2}+0.28\pi r$

$C'(r)=0$

$-\frac{8}{r^2}+0.28\pi r=0$

$0.28\pi r=\frac{8}{r^2}$

$r^3=\frac{8}{0.28\pi}=9.095$

$r=(9.095)^{\frac{1}{3}}=2.09$

Again, differentiate w.r.t r

$C''(r)=\frac{16}{r^3}+0.28\pi$

Substitute the value of r

$C''(2.09)=\frac{16}{(2.09)^3}+0.28\pi=2.63>0$

Therefore,the product cost is minimum at r=2.09

h= $\frac{200}{\pi (2.09)^2}=14.57$

Radius of can,r=2.09 cm

Height of cone,h=14.57 cm

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3 years ago

Help please I will be marking brainliest!!!

viktelen [127]

Answer:

Step-by-step explanation:

d

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