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2 years ago

Find the volume

Mathematics

2 answers:

bazaltina [42]2 years ago

5 0

Answer:

808.5 m³

Step-by-step explanation:

Volume of a Triangular Prism :

Volume = 1/2 × b × h × l

=============================================================

Given :

⇒ b = 11 m

⇒ h = 10.5 m

⇒ l = 14 m

===========================================================

Solving :

⇒ V = 1/2 × 11 × 10.5 × 14

⇒ V = 7 × 115.5

⇒ V = 808.5 m³

PSYCHO15rus [73]2 years ago

3 0

Area of base

1/2BH
1/2(10.5)(14)
1/2(147)
73.5m²

Volume

Area of base×Height

73.5(11)
808.5m³

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Answer:

Company C

Step-by-step explanation:

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3 years ago

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two dice are rolled in succession. determine the probability that their sum is 10 or more,given that the first die is a 5

MrRa [10]

If the first die is a 5, the second must be 5 or more in order to have a sum of at least ten.

Assuming it is a six sided die, the two values that would make this true are 5 and 6.

2 values/6 possible would be a probability of 2/6, which simplifies to 1/3

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3 years ago

Assume that the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder. Based on this assumption,

kompoz [17]

If the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder, then its volume is

$V_{flask}=V_{sphere}+V_{cylinder}.$

Use following formulas to determine volumes of sphere and cylinder:

$V_{sphere}=\dfrac{4}{3}\pi R^3,\\ \\V_{cylinder}=\pi r^2h,$

wher R is sphere's radius, r - radius of cylinder's base and h - height of cylinder.

Then

$V_{sphere}=\dfrac{4}{3}\pi R^3=\dfrac{4}{3}\pi \left(\dfrac{4.5}{2}\right)^3=\dfrac{4}{3}\pi \left(\dfrac{9}{4}\right)^3=\dfrac{243\pi}{16}\approx 47.71;$
$V_{cylinder}=\pi r^2h=\pi \cdot \left(\dfrac{1}{2}\right)^2\cdot 3=\dfrac{3\pi}{4}\approx 2.36;$
$V_{flask}=V_{sphere}+V_{cylinder}\approx 47.71+2.36=50.07.$

Answer 1: correct choice is C.

If both the sphere and the cylinder are dilated by a scale factor of 2, then all dimensions of the sphere and the cylinder are dilated by a scale factor of 2. So

R'=2R, r'=2r, h'=2h.

Write the new fask volume:

$V_{\text{new flask}}=V_{\text{new sphere}}+V_{\text{new cylinder}}=\dfrac{4}{3}\pi R'^3+\pi r'^2h'=\dfrac{4}{3}\pi (2R)^3+\pi (2r)^2\cdot 2h=\dfrac{4}{3}\pi 8R^3+\pi \cdot 4r^2\cdot 2h=8\left(\dfrac{4}{3}\pi R^3+\pi r^2h\right)=8V_{flask}.$