A number cube with the numbers 1 through 6 is rolled. Find the probability of rolling the number 4.

Airida [17]

4.415e+8

Have a wonderful day

8 0

3 years ago

A pencil consists of a cone stacked on top of a cylinder. The diameter of the cylindrical base of the pencil is 10 mm and the he

11Alexandr11 [23.1K]

Answer:

790π

Step-by-step explanation:

We are given;

Diameter of cylinder;d = 10 mm

So, radius;r = 10/2 = 5 mm

Height of cylinder;h = 70mm

Surface area of cylinder is given by the formula; S.A = 2πr² + 2πrh

Plugging in the relevant values, we have;S.A = 2π(5)² + 2π(5)(70)

S.A = 50π + 700π

S.A = 750π

Now, because one base of the cylinder is hidden as the cone is stacked on that face, we will deduct the area of that base face;

Thus, Surface area = 750π - π(5)² = 750π - 25π = 725π

For the cone,

Height;h = 12mm

Since this is stacked directly on the cylinder, it will have the same radius. Thus; radius;r = 5mm

Now,formula for surface area of cone is;

S.A = πr² + πrL

Where L is slant height.

We can use pythagoras theorem to get L.

So, L² = r² + h²

L = √r² + h²

L = √(5² + 12²)

L = √(25 + 144)

L = √169

L = 13

So, S.A of cone = π(5)² + (π×5×13)

S.A = 25π + 65π = 90π

Similar to what was done to the Cylinder, since the circular base of the cone is stacked on the cylinder, we will deduct the surface area of that base as it is hidden.

So, S.A is now = 90π - π(5)²

= 90π - 25π = 65π

Thus,total surface area of the pencil = 725π + 65π = 790π

7 0

3 years ago

Read 2 more answers

A spider has 8 legs. An ant has 6 legs. There are a group of spiders and a group of ants. The groups have equal numbers of legs.

Lemur [1.5K]

There are three spiders and 4 ants in all

8 0

3 years ago

Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$ .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

$y-y_1=m(x-x_1)$

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0

3 years ago

HELP!!!!!! What is the equation of the axis of symmetry for the parabola

Dima020 [189]

Answer:

The line of symmetry is x = 3

Step-by-step explanation:

To find the line of symmetry in this case, use the base formula for vertex form.

y = a(x - h) + k

In this formula, h is the line of symmetry. In this case, h is 3.

3 0

3 years ago