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IrinaK [193]
3 years ago
8

about 7.5 gallons of water will fill up 1 cubic foot of space. how many gallons of water will fill a goldfish pool shaped like t

he prism shown
Mathematics
2 answers:
Cloud [144]3 years ago
7 0
Well, since there is no image I am unable to say the exact volume. But here is the volume formula-
V=L times W times H
*V=volume W=width H=height
and you don't actually have to write times you can do the symbol.

Example- if I had a rectangular prism of the width being 5 cubes, the length 8 cubes, and the height 7 cubes(following how many gallons are in a cube 7.5) I would the use my formula. V=LWH- 8 times 7= 56 56 times 5=280 280 times 7.5= 2100
V=2100 (that is your answer reply If you didn't mean the volume or you don't get it)


bekas [8.4K]3 years ago
6 0

Hi, I don't really know if this is correct but oh well.

<u>Answer:</u>

11 5/6

<u>Step-By-Step Explaination:</u>

V = l·w·h

3 1/2·7.5 = 5 1/4

5 1/4 · 3 = 15 3/4

15 3/4 · 7.5 = 11 5/6

Hope this helped. :)

(just noticed this it wrong sorry)

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About 80% of e-waste is shipped outside of the US. Which of the following statements about e-waste is false? A) E-waste is routi
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Step-by-step explanation:

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3 years ago
A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of th
Leno4ka [110]

The value of the angle subtended by the distance of the buoy from the

port is given by sine and cosine rule.

  • The bearing of the buoy from the is approximately <u>307.35°</u>

Reasons:

Location from which the ship sails = Port

The speed of the ship = 10 mph

Direction of the ship = N35°W

Location of the warning buoy = 5 miles north of the port

Required: The bearing of the warning buoy from the ship after 7.5 hours.

Solution:

The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles

By cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

Where;

a = The distance between the ship and the buoy

b = The distance between the ship and the port = 75 miles

c = The distance between the buoy and the port = 5 miles

Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°

Which gives;

a = √(75² + 5² - 2 × 75 × 5 × cos(35°))

By sine rule, we have;

\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}

Therefore;

\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}

Which gives;

\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}

\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}

Similarly, we can get;

\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}

The angle subtended by the distance of the buoy from the port, <em>C</em> is therefore;

C ≈ 180° - 142.68° - 35° ≈ 2.32°

By alternate interior angles, we have;

The bearing of the warning buoy as seen from the ship is therefore;

Bearing of buoy ≈ 270° + 35° + 2.32° ≈ <u>307.35°</u>

Learn more about bearing in mathematics here:

brainly.com/question/23427938

5 0
2 years ago
A boy’s age is 5 years less than one- third of his father’s age. If the difference in their age is 23, what is the age of the bo
elena-14-01-66 [18.8K]
6

23-5=18
18/3=6

Hope it helps
6 0
1 year ago
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Help!!<br><br> State two ways that you could use a slope of -5/3 to graph a line.
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Answer:

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Step-by-step explanation:

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3 how far you move left on the graph

7 0
3 years ago
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