What is (13x + 9k) + (17x + 6k), simplified?

A gum ball has a radius that is 30mm. The radius of the gumballs spherical hollow core is 28mm. What is the volume of the gum ba

Arada [10]

Answer:

V = 21145.01 mm³

Step-by-step explanation:

Given that:

The radius of the gumball = 30 mm

The radius of the spherical hollow-core = 28 mm

Consider the radius of the gumball to be $r_2$ and the radius of the spherical hollow-core to be $r_1$ . Then;

The volume of the gumball can be determined by using the formula of a sphere.

$V = \dfrac{4}{3}\pi (r_2^3 - r_1^3)$

$V = \dfrac{4}{3} \times \pi \times (30^3 - 28^3)$

V = 21145.01 mm³

5 0

3 years ago

The equation for a circle is x^2 − 8x + y^2 - 2y - 8 = 0

Ivanshal [37]

Answer:

B

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

To obtain this form use the method of completing the square

Given

x² - 8x + y² - 2y - 8 = 0 ( add 8 to both sides )

x² - 8x + y² - 2y = 8

add ( half the coefficient of the x/y terms )² to both sides

x² + 2(- 4)x + 16 + y² + 2(- 1)y + 1 = 8 + 16 + 1

(x - 4)² + (y - 1)² = 25 → B

6 0

2 years ago

A plane flying horizontally at an altitude of 3 miles and a speed of 500 mi/h passes directly over a radar station. Find the rat

konstantin123 [22]

Answer:

The rate at which the distance from the plane to the station is increasing is 331 miles per hour.

Step-by-step explanation:

We can find the rate at which the distance from the plane to the station is increasing by imaging the formation of a right triangle with the following dimensions:

a: is one side of the triangle = altitude of the plane = 3 miles

b: is the other side of the triangle = the distance traveled by the plane when it is 4 miles away from the station and an altitude of 3 miles

h: is the hypotenuse of the triangle = distance between the plane and the station = 4 miles

First, we need to find b:

$a^{2} + b^{2} = h^{2}$ (1)

$b = \sqrt{h^{2} - a^{2}} = \sqrt{(4 mi)^{2} - (3 mi)^{2}} = \sqrt{7} miles$

Now, to find the rate we need to find the derivative of equation (1) with respect to time:

$\frac{d}{dt}(a^{2}) + \frac{d}{dt}(b^{2}) = \frac{d}{dt}(h^{2})$

$2a\frac{da}{dt} + 2b\frac{db}{dt} = 2h\frac{dh}{dt}$

Since "da/dt" is constant (the altitude of the plane does not change with time), we have:

$0 + 2b\frac{db}{dt} = 2h\frac{dh}{dt}$

And knowing that the plane is moving at a speed of 500 mi/h (db/dt):

$\sqrt{7} mi*500 mi/h = 4 mi*\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{\sqrt{7} mi*500 mi/h}{4 mi} = 331 mi/h$

Therefore, the rate at which the distance from the plane to the station is increasing is 331 miles per hour.

I hope it helps you!

4 0

2 years ago

You have $40.00. You wish to buy a T-shirt costing $14.50. You would also like to buy a pair of jeans. There

atroni [7]

Tax is 1.54

Including tax = 39.99
Excluding tax= 38.5

Hope this helps

3 0

1 year ago

Number 8 please thanks!

WINSTONCH [101]

$4b(b+2)=16+4b^2\\4b^2+8b=16+4b^2\\8b = 16 + 4b^2-4b^2\\8b = 16\\b = 2$

Hope that helps!

4 0

2 years ago