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1 year ago

Example 4

Mathematics

1 answer:

Vesnalui [34]1 year ago

7 0

Answer: Part a: b represents the base fee charged for the "house call" job

m is the hourly rate.

x is the number of hours at the job.

y represents the total cost for the job.

Part b:. The y-intercept is 65

The equation follows y=mx+b

Y= 42x + 65

Part c:. $275

Step-by-step explanation:

The difference between the two total costs, (Camilla 191 - Connor 107)

= $84. That is the rise. The run is the difference in time, 2 hours. 84/2 = 42.

m= 42

42 from 107 to find b, the base fee, 65 This is the y-intercept.

The equation is y= 42x + 65

Quick check using Camilla's bill: 191 = (42)(3) + 65

191 = 126 + 65 True!

Part C:

To find the total cost of the plumber's 5 hour job,

Y = 42(5) + 65

Y= 210 + 65

Y= $275

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Answer:

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Step-by-step explanation:

The complete question in the attached figure

we know that

When you get to the second checkpoint, you are 1/4 of the distance to the finish

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The distance from the star to the second checkpoint is

$\frac{1}{4}(40)=10\ miles$

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2 years ago

Jorden drives to the store at 30 miles per hour. On her way home she averages only 20 miles per hour. If the total driving time

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For this case we have the following definition:
$d = v * t
$
Where,
d: distance
v: speed
t: time
Clearing the time we have:
$t = d / v
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The total driving time takes half an hour:
$t = d / v1 + d / v2

1.5 = d / 30 + d / 20$
Clearing d we have:
$1.5 * 30 * 20 = 20d + 30d

1.5 * 30 * 20 = 50d

d = (1.5 * 30 * 20) / (50)

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2 years ago

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

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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

We can conclude that there are 12 faces in on the four geometric shapes on the holes.

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.

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2 years ago

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Answer:

Step-by-step explanation:

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1 year ago

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hodyreva [135]

The best correct answer is option E. If two or more polygons are congruent, then rigid transformations can be used to map one polygon onto the other. Two figures are said to be congruent when they have the same shape and size or if one object is a mirror image of the other object.

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