Help ASAP please

Find the volume of the composite figure. Explain your thought, explain how your arrived at your answer. SHOW YOUR WORK FOR FULL

Bond [772]

At the bottom of the composite figure we have half a sphere, of radius 10 in.

The volume of this hemisphere would be half the volume of the full sphere, or:

(1/2)(4/3)π(10 in)^3, or (2/3)π(1000 in^3), or (2000/3)π in^3.

On top is the cone of radius 10 and slant height 15 in. To find the volume of this cone-shaped solid, we'll need the height of the cone. This can be found using the Pyth. Thm. as follows:

15^2 = 10^2 + h^2, where h is the height of the cone.

225 = 100 + h^2, so that h= √125, or 5√5. The height of the cone is 5√5 in.

Then the volume of the cone is V = (1/3)(base)(height)

= (1/3)(π)(100 in^2)(5√5 in)

= 500√5/3(π) in^3

The total volume of the composite solid is then

(2000/3)(π in^3) + ( 500√5/3(π) ) in^3), or

(π/3)(4+√5) in^3. This comes out to 6.53 in^3, to the nearest hundredth.

7 0

2 years ago

How do i find the answer to4k+mn=n-3 it's a literal equation formula and i have to solve for n

Bezzdna [24]

4k+mn=n-3
3(4k+mn)= 12k+mn
So, 12k+mn=n

7 0

2 years ago

A child's sandbox is 5 feet wide, 4 feet long, and 3 feet deep. Grace fills the sandbox so that the sand is 2 ½ feet deep. What

8_murik_8 [283]

Answer:

the volume of sand is $50 ft^{3}$

Step-by-step explanation:

This problem bothers on the mensuration of solid shapes, rectangular prism.

Given data

$Length l= 4 ft Width w= 5ft\\Height h = 3ft$

if the sandbox is filled with sand of $2\frac{1}{2 } ft$ deep the volume of the sand in the box will be calculated based on the the depth of the sand

converting the depth of sand from mixed fraction to proper fraction we have $2\frac{1}{2} = \frac{5}{2}$

the expression for the volume of a rectangular prism is

$volume of sand = length * width* height$

substituting our data into the expression we have

$volume of sand= 5*4 * \frac{5}{2} \\volume of sand = \frac{100}{2} \\volume of sand = 50 ft^{3}$

5 0

2 years ago

WILL MARK BRAINLIEST!!!!!

statuscvo [17]

The shape with a rotational symmetry would remain the same when rotated

The shape with the smallest angle of rotational symmetry of 180 degrees is shape D

<h3>How to determine the shape?</h3>

To do this, we simply examine each of the options

<u>Shape A: Cross</u>

The shape has 4 equal sides.

So, the smallest angle of rotational symmetry is:

Angle = 360/4

Angle = 90

The smallest angle of rotational symmetry is 90 degrees

<u>Shape B: Star</u>

The shape has 5 vertices.

So, the smallest angle of rotational symmetry is:

Angle = 360/5

Angle = 72

The smallest angle of rotational symmetry is 72 degrees

<u>Shape C: Composite figure</u>

This shape has a square and a triangle merged together at two ends

This means that the smallest angle of rotational symmetry is a complete rotation of 360 degrees

<u>Shape D: Circle</u>

The circle is divided into two equal segments.

The angle on the straight line of each segment is:

Angle = 360/2

Angle = 180

The smallest angle of rotational symmetry is 180 degrees

Hence, the shape with the smallest angle of rotational symmetry of 180 degrees is shape D

Read more about rotational symmetry at:

brainly.com/question/12214455

8 0

2 years ago

A. Solve the differential equation <img src="https://tex.z-dn.net/?f=y%27%3D2x%20%5Csqrt%7B1-y%5E2%7D%20" id="TexFormula1" title

kirill [66]

$y' = \frac{dy}{dx}$

seperable differential equations will have the form
$\frac{dy}{dx} = F(x) G(y)$

what you do from here is isolate all the y terms on one side and all the X terms on the other
$\frac{dy}{G(y)} = F(x) dx$
just divided G(y) to both sides and multiply dx to both sides

then integrate both sides
$\int \frac{1}{G(y)} dy = \int F(x) dx

$

once you integrate, you will have a constant. use the initial value condition to solve for the constant, then try to isolate x or y if the question asks for it

In your problem,
$G(y) = \sqrt{1-y^2}

F(x) = 2x$

so all you need to integrate is
$\int \frac{1}{\sqrt{1-y^2}} dy = \int 2x dx$

5 0

3 years ago