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3 years ago

If the area of the circle below is 18 ft2, what is the area of the shaded sector?

Mathematics

2 answers:

adoni [48]3 years ago

8 0

Answer:

C.4.5 ft^2

Step-by-step explanation:

If the area of the whole circle is 18 ft^2, and a whole circle is 360 degrees, and you are looking for the area of 90 degrees of that circle (which is one fourth of the circle) you would divide the area by 4, to get the area for that sector.

360/90= 4: meaning that 90 degrees is 1/4 of a circle.

So, you would divide 18 by 4.

18/4=4.5

Hope this helps!

LuckyWell [14K]3 years ago

5 0

Answer:

It would be C

Step-by-step explanation:

since the whole area of the circle is 18 and the sector takes up one-fourth of the circle you would just divide 18 by 4 to get 4.5

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A toy rocket is launched from a platform that is 48 ft high. The rocket height above the ground is modeled by h=-16t^2+32t+48. a

Komok [63]

Answer:

a) 64 feet

b) 3 seconds

Step-by-step explanation:

The maximum height of $h=h(t)$ can be bound by finding the y-coordinate of the vertex of $y=-16x^2+32x+48$ .

Compare this equation to $y=ax^2+bx+c$ to find the values of $a,b,\text{ and } c$ .

$a=-16$

$b=32$

$c=48$ .

The x-coordinate of the vertex can be found by evaluating:

$\frac{-b}{2a}=\frac{-32}{2(-16)$

$\frac{-b}{2a}=\frac{-32}{-32}$

$\frac{-b}{2a}=1$

So the x-coordinate of the vertex is 1.

The y-coordinate can be found be evaluating $y=-16x^2+32x+48$ at $x=1$ :

$y=-16(1)^2+32(1)+48$

$y=-16+32+48$

$y=16+48$

$y=64$

So the maximum height of the rocket is 64 ft high.

When the rocket hit's the ground the height that the rocket will be from the ground is 0 ft.

So we are trying to find the second t such that:

$0=-16t^2+32t+48$

I'm going to divide both sides by -16:

$0=t^2-2t-3$

Now we need to find two numbers that multiply to be -3 and add to be -2.

Those numbers are -3 and 1 since (-3)(1)=-3 and (-3)+(1)=-2.

$0=(t-3)(t+1)$

This implies we have either $t-3=0$ or $t+1=0$

The first equation can be solved by adding 3 on both sides: $t=3$ .

The second equation can be solved by subtracting 1 on both sides: $t=-1$ .

So when $t=3$ seconds, is when the rocket has hit the ground.

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3 years ago

A circular cylindrical container, open at the top, and having a capacity of 24pi cubic inches, is to be manufactured. If the cos

otez555 [7]

Answer: radius r = 2 inches

height h = 6 inches

Step-by-step explanation:

Given;

Volume V = 24πin3

Volume of a cylinder is given by

V = πr^2h

h = V/πr^2. ....1

Where, h = height and r = radius of cylinder

For the surface area of the cylinder with open top. we have,

S = 2πrh + πr^2

For the cost of materials used, let k represent the cost of materials used for the body of the cylinder.

Then, for the bottom will be 3k

Total cost will be represented by C, which gives

C = 2πrhk + 3πr^2k. .....2

Substituting eqn 1 to 2, we have;

C = 2πrVk/πr^2 + 3πr^2k

C = 2Vk/r + 3πr^2k

The material cost is minimum at dC/dr = 0

dC/dr = -2Vk/r^2 + 6πrk =0

6πrk = 2Vk/r^2

r^3 = 2V/6π

r = (2×24π/6π)^-3

r = (8)^-3

r = 2

Substituting r = 2 into eqn1

h = 24π/π(2^2)

h = 24/4 = 6

h = 6

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-BARSIC- [3]

Answer:

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y-intercept - y = 2x + 6.3

Step-by-step explanation:

For x,

-6x + 3y = 18.9

-6x = -3y + 18.9

x = -3y/-6 + 18.9/-6

x = 1/2y - 3.15

For y,

-6x + 3y = 18.9

3y = 6x + 18.9

y = 6x/3 + 18.9/3

y = 2x + 6.3

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