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dedylja [7]
3 years ago
14

How do I find the volume of this right triangular prism?

Mathematics
1 answer:
frozen [14]3 years ago
8 0
I’ll try to make a step by step demonstration! Feel free to skip to step 5 for the answer, though
1. The formula for a right triangular prism is (1/2)*base*height*length, and we know what the base (4 cm) and length (8 cm) are, but not the height
2. To find the height we must divide the side triangle into two, which then gives us two right triangles
3. Focusing on just one of the right triangles, its base will be 2 cm rather than 4 (as it has been divided by two) and its hypotenuse will be 4 cm; using Pythagorean’s theorem (a^2 + b^2 = c^2), solve for the missing side (the height), which should end up being √12
4. Multiply! (1/2)4*√12*8 ≈ 55.425
5. Round to the nearest tenth —> 55.4
I hope this helped somewhat! Comment if clarification is needed
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7 = y

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5 0
3 years ago
0.12y-1=0.095-0.9 <br><br>help
balu736 [363]

0.12y-1=0.095-0.9

or, 0.12y=0.095-0.9+1

or, 0.12y=0.195

or, y = 0.195/0.12 = 195/120=1.625

The answer will be 1.625

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3 years ago
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NEED HELP QUICK! Points P, Q, and R are shown on the number line. What is the distance between point P and point R? A) 47 units
FromTheMoon [43]
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A car is changing speed from 11 m/s to 26 m/s in a time interval of 3.0 s.
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7 0
4 years ago
Which hyperbola has one focus in common with the hyperbola x^2/16 - y^2/9 = 1
Goshia [24]

Answer:

The same focus is (-5 , 0) ⇒ Answer D

Step-by-step explanation:

* Lets study the equation of the hyperbola

# The standard form of the equation of a hyperbola with  

  center (0 , 0) and transverse axis parallel to the x-axis is

  x²/a² - y²/b² = 1

- The coordinates of the foci are (± c , 0),  where c² = a² + b²

# The standard form of the equation of a hyperbola with  

  center (h , k) and transverse axis parallel to the x-axis is

  (x - h)²/a² - (y - k)²/b² = 1

- the coordinates of the foci are (h ± c , k), where c² = a² + b²

# The standard form of the equation of a hyperbola with  

  center (h , k) and transverse axis parallel to the y-axis is

  (y - k)²/a² - (x - h)²/b² = 1

- the coordinates of the foci are (h , k ± c), where c² = a² + b²

* Now lets solve the problem

∵ x²/16 - y²/9 = 1

∴ a² = 16 and b² = 9

∵ c² = a² + b²

∴ c² = 16 + 9 = 25 ⇒ take √ to find the values of c

∴ c = ±√25 = ± 5

∴ The foci are (5 , 0) , (-5 , 0)

# Answer A:

∵ (y - 5)/16 - (x - 13)/9 = 1

∵  (y - k)²/a² - (x - h)²/b² = 1

∴ The foci are (h , k + c) , (h , k - c)

∴ h = 13 and k = 5

∵ a² = 16 and b² = 9

∵ c² = a² + b²

∴ c² = 16 + 9 = 25 ⇒ take √ to find the values of c

∴ c = ±√25 = ± 5

∴ The foci are (13 , 5+5) , (13 , 5-5)

∴ The foci are (13 , 10) , (13 , 0) ⇒ not the same

# Answer B:

∵ (x - 13)²/25 - (y - 5)²/144

∵ (x - h)²/a² - (y - k)²/b² = 1

∵ The foci are (h ± c , k)

∴ h = 13 and k = 5

∵ a² = 25 and b² = 144

∵ c² = a² + b²

∴ c² = 125 + 144 = 169 ⇒ take √ to find the values of c

∴ c = ±√169 = ± 13

∴ The foci are (13 + 13 , 5) , (13 - 13 , 5)

∴ The foci are (26 , 5) , (0 , 5) ⇒ not the same

# Answer C:

∵ (y - 5)/25 - (x - 13)/144 = 1

∵  (y - k)²/a² - (x - h)²/b² = 1

∴ The foci are (h , k + c) , (h , k - c)

∴ h = 13 and k = 5

∵ a² = 25 and b² = 144

∵ c² = a² + b²

∴ c² = 25 + 144 = 169 ⇒ take √ to find the values of c

∴ c = ±√169 = ± 13

∴ The foci are (13 , 5+13) , (13 , 5-13)

∴ The foci are (13 , 18) , (13 , -8) ⇒ not the same

# Answer D:

∵ (y + 13)/144 - (x + 5)/25 = 1

∵  (y - k)²/a² - (x - h)²/b² = 1

∴ The foci are (h , k + c) , (h , k - c)

∴ h = -5 and k = -13

∵ a² = 144 and b² = 25

∵ c² = a² + b²

∴ c² = 144 + 25 = 169 ⇒ take √ to find the values of c

∴ c = ±√169 = ± 13

∴ The foci are (-5 , -13+13) , (-5 , -13-13)

∴ The foci are (-5 , 0) , (-5 , -26) ⇒ one of them the same

* The same focus is (-5 , 0)

8 0
4 years ago
Read 2 more answers
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