Observe the graph below.<br> The graph shows a<br> function that is asymptotic to

You are designing a miniature golf course and need to calculate the surface area and volume of many of the objects that will be

laiz [17]

a. To solve the first part, we are going to use the formula for the surface area of a sphere: $A=4 \pi r^2$
where
$A$ is the surface area of the sphere
$r$ is the radius of the sphere
We know from our problem that $r=5ft$ ; so lets replace that value in our formula:
$A=4 \pi (5ft)^2$
$A=314.16ft^2$

To solve the second part, we are going to use the formula for the volume of a sphere: $V= \frac{4}{3} \pi r^3$
Where
$V$ is the volume of the sphere
$r$ is the radius
We know form our problem that $r=5ft$ , so lets replace that in our formula:
$V= \frac{4}{3} \pi (5ft)^3$
$V=523.6ft^3$

We can conclude that the surface area of the sphere is 314.16 square feet and its volume is 523.6 cubic feet.

b. To solve the first part, we are going to use the formula for the surface area of a square pyramid: $A=a^2+2a \sqrt{ \frac{a^2}{4} +h^2}$
where
$A$ is the surface area
$a$ is the measure of the base
$h$ is the height of the pyramid
We know form our problem that $a=8ft$ and $h=12ft$ , so lets replace those value sin our formula:
$A=(8ft)^2+2(8ft) \sqrt{ \frac{(8ft)^2}{4} +(12ft)^2}$
$A=266.39ft^2$

To solve the second part, we are going to use the formula for the volume of a square pyramid: $V=a^2 \frac{h}{3}$
where
$V$ is the volume
$a$ is the measure of the base
$h$ is the height of the pyramid
We know form our problem that $a=8ft$ and $h=12ft$ , so lets replace those value sin our formula:
$V=(8ft)^2 \frac{(12ft)}{3}$
$V=256ft^3$

We can conclude that the surface area of our pyramid is 266.39 square feet and its volume is 256 cubic feet.

c. To solve the first part, we are going to use the formula for the surface area of a circular cone: $A= \pi r(r+ \sqrt{h^2+r^2}$
where
$A$ is the surface area
$r$ is the radius of the circular base
$h$ is the height of the cone
We know form our problem that $r=5ft$ and $h=8ft$ , so lets replace those values in our formula:
$A= \pi (5ft)[(5ft)+ \sqrt{(8ft)^2+(5ft)^2}]$
$A=226.73ft^2$

To solve the second part, we are going to use the formula for the volume os a circular cone: $V= \pi r^2 \frac{h}{3}$
where
$V$ is the volume
$r$ is the radius of the circular base
$h$ is the height of the cone
We know form our problem that $r=5ft$ and $h=8ft$ , so lets replace those values in our formula:
$V= \pi (5ft)^2 \frac{(8ft)}{3}$
$V=209.44ft^3$

We can conclude that the surface area of our cone is 226.73 square feet and its surface area is 209.44 cubic feet.

d. To solve the first part, we are going to use the formula for the surface area of a rectangular prism: $A=2(wl+hl+hw)$
where
$A$ is the surface area
$w$ is the width
$l$ is the length
$h$ is the height
We know from our problem that $w=6ft$ , $l=10ft$ , and $h=16ft$ , so lets replace those values in our formula:
$A=2[(6ft)(10ft)+(16ft)(10ft)+(16ft)(6ft)]$
$A=632ft^2$

To solve the second part, we are going to use the formula for the volume of a rectangular prism: $V=whl$
where
$V$ is the volume
$w$ is the width
$l$ is the length
$h$ is the height
We know from our problem that $w=6ft$ , $l=10ft$ , and $h=16ft$ , so lets replace those values in our formula:
$V=(6ft)(16ft)(10ft)$
$V=960ft^3$

We can conclude that the surface area of our solid is 632 square feet and its volume is 960 cubic feet.

e. Remember that a face of a polygon is a side of polygon.
- A sphere has no faces.
- A square pyramid has 5 faces.
- A cone has 1 face.
- A rectangular prism has 6 faces.
Total faces: 5 + 1 + 6 = 12 faces

<span>We can conclude that there are 12 faces in on the four geometric shapes on the holes.
</span>
f. Remember that an edge is a line segment on the boundary of the polygon.
- A sphere has no edges.
- A cone has no edges.
- A rectangular pyramid has 8 edges.
- A rectangular prism has 12 edges.
Total edges: 8 + 20 = 20 edges

Since we have 20 edges in total, we can conclude that your boss will need 20 brackets on the four shapes.

g. Remember that the vertices are the corner points of a polygon.
- A sphere has no vertices.
- A cone has no vertices.
- A rectangular pyramid has 5 vertices.
- A rectangular prism has 8 vertices.
Total vertices: 5 + 8 = 13 vertices

We can conclude that there are 0 vertices for the sphere and the cone; there are 5 vertices for the pyramid, and there are are 8 vertices for the solid (rectangular prism). We can also conclude that your boss will need 13 brackets for the vertices of the four figures.

7 0

2 years ago

489 to the nearest ten

sammy [17]

8 is in the tenth position.

In order to round up, the number behind it (in this case. 9) must be one of the numbers of 5-9. Because 9 meets this requirement, you can round 8 to 9 and this will make 9 to 0.

Your answer is 490

4 0

2 years ago

Read 2 more answers

The perimeter of the school crossing sign is 102 inches what is the length of each side

Afina-wow [57]

The perimeter of the school crossing sign is 102 inches
length of each side = ?
in the question number of sides is not mentioned, so the answer may vary for the length of each side. if there is 5 sides divide 102 by 5, we get 20.4 answer so in case of 5 sides length of each side is 20.4 inches. if there is 4 sides divide 102 by 4 and we get 25.5 inches answer for the legnth each side.

6 0

2 years ago

The student to faculty ratio at a small college is 17:3 the total number of students and faculty is 740. How many faculty member

larisa86 [58]

Answer:

111

Step-by-step explanation:

It is given that the student to faculty ratio at a small college is 17:3.

Let number of students and faculty members are 17x and 3x respectively.

It is given that total number of students and faculty is 740.

$17x+3x=740$

$20x=740$

Divide both sides by 20.

$x=37$

Number of student $=17\times 37=629$

Number of faculty members $=3\times 37=111$

Therefore, there are 111 faculty members at the college.

3 0

2 years ago

Quadrilateral ABCD is similiar to quadrilateral EFGH. The lengths of the three longest sides in quadrilateral ABCD are 24 feet,

masha68 [24]

Well the sides of these 2 quadrilaterals are proportional to each other , so you would have to find the scale factor of the 2 then see which sides have already proportional pairs , then multiply *or / to find the last one .

Answer: D. 11.7 ft

5 0

2 years ago

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Observe the graph below. The graph shows a function that is asymptotic to