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ludmilkaskok [199]
2 years ago
11

Acapulco, Mexico and Hyderabad, India both lie at 17° north latitude, and lie very nearly halfway around the world from each oth

er in an east-west direction. The radius of Earth at a latitude of 17° is about 3790 miles. Suppose that you could fly from Acapulco directly west to Hyderabad or fly directly north to Hyderabad. Which way would be shorter, and by how much? Use 3960 miles for Earth’s radius. (Hint: To fly directly north, you would go from 17° north latitude to 90° north latitude, and then back down to 17° north latitude.)
Mathematics
1 answer:
Nataly_w [17]2 years ago
6 0
In solving this problem we can consider Earth to be a sphere. When we have a circle, then we can use this formula to find arc length:
L=2 \pi R \frac{C}{360}
Where:
L= arc length (in this problem it is disance we need to travel)
R = radius of circle (in this problem it is equal to a radius of Earth)
C = angle we need to pass

We are told that two cities lie halfway around the world. If we fly to west this means angle is 180°.
This gives an arc length of:
L=2* \pi *3960* \frac{180}{360}  \\  \\ L=12440.71 miles

If we want to fly to north we need to go to 90° northern latidtude and then back to 17° latitude. This means angle is:
C=2*(90-17)=2*73°=146°
This gives an arc length of:
L=2* \pi *3960* \frac{146}{360} \\ \\ L=10090.8 miles

We can see that flying north is shorter. It is shorter by:
12440.71 miles - 10090.8 miles = 2349.91 miles
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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
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
The area of a rectangular painting is given by the trinomial x2 + 4x - 21. What are the possible dimensions of the painting? Use
zloy xaker [14]
To find the dimensions you can just factor the equation...

x^2+4x-21

x^2-3x+7x-21

x(x-3)+7(x-3)

(x+7)(x-3) and this is equal to LW or WL

And note that x>3 for any possible solution.

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2 years ago
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A dog has dug holes in diagonally-opposite corners of a rectangular yard. One length of the yard is 8 meters and the distance be
LUCKY_DIMON [66]

For a better understanding of the solution provided please go through the diagram in the file attached.

Let ABCD be the rectangular yard. The diagonal d=17 meters. AD=8 meters. Therefore, the length of DC can be found by applying the Pythagorean theorem in the right triangle \Delta ADC as:

DC=\sqrt{17^2-8^2}=\sqrt{(AC)^2-(AD)^2}=\sqrt{225} =15 meters.

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2 years ago
Gabriel makes a model of a pyramid with the dimensions shown. A square pyramid. The square base has side lengths of 12 inches. T
Blizzard [7]

Answer:

The area of the square base is 114 in. 2

The area of each triangular face is 66 in. 2

Gabriel will need 408 in. 2  of paint

8 0
2 years ago
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Faelyn grouped the terms and factored the GCF out of the groups of the polynomial 6x4 – 8x2 + 3x2 + 4. Her work is shown.
Y_Kistochka [10]
It is hard to say how Faelyn's work shows the polynomial is prime, but it is.

There are no factors of 6*4 = 24 that add to -5, so the polynomial cannot be factored using real numbers.
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The expression of suggestion 2 is a different polynomial than the one Faelyn is factoring.
Taking a factor of 2x out of the first group does not help it match the factoring of the second group.
Dividing one or the other of the groups by -1 will not make the binomials the same.
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