Cones A and B both have volume of 48LaTeX: \pi π cubic units, but have different dimensions. Cone A has radius (r) 6 units and height (h) 4 units. Select a possible radius (r) and height (h) for Cone B.

Radius = 2cm ; height = 36cm

Step-by-step explanation:

Recall:

Volume of a cone (V):

1/3πr²h

r = radius ; h = height

Given that:

Cone A:

Va = 48π

ha = 4 ; ra = 6

Cone B :

Vb = 48π

hb = ? ; rb =?

Since both cones have similar volume:

Va = Vb

1/3πra²ha = 1/3πrb²hb

48π = 1/3πrb²hb

48 = 1/3rb²hb

48*3 = rb²hb

144 = rb²hb

One possible solution is :

144 = 36 * 4

Where ;

r² =4

h = 36

r = √4 ; r = 2

h = 36

Radius = 2cm ; height = 36cm

y = 2x

step-by-step explanation:

if you divide each number in the sequence, you'll see that they all have a multiple factor of 2. each number in the series, and any other numbers which may continue, will be multiplied by 2. therefore, y is the next number in the sequence, and x is the current one, so, our formula will be y = 2x.

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204