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Ilia_Sergeevich [38]
3 years ago
13

Which solid figure has a base that is a polygon and triangular faces that meet

Mathematics
2 answers:
sladkih [1.3K]3 years ago
6 0
The answer is c. prism
Lady bird [3.3K]3 years ago
3 0

Answer:

B. Pyramid

Step-by-step explanation:

In the beginning, we can eliminate C and D because they don't have any triangular faces. A cone has a circle as a base. A circle is not a polygon. So, B is the answer because it has triangular faces <em>and</em> a square (polygon) base.

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In the drawing above two rugby players start 41 meters apart at rest. Player 1 accelerates at 2.2 m/s^2 and player 2 accelerates
Anettt [7]

Answer:

The time that elapses before the players collide is 4.59 secs

Step-by-step explanation:

The distance between the two players is 41 meters. Hence, they would have total distance of 41 meters by the time they collide.

We can write that

S₁ + S₂ = 41 m

Where S₁ is the distance covered by Player 1 before collision

and S₂ is the distance covered by Player 2 before collision

From one of the equations of kinematics for linear motion,

S = ut + \frac{1}{2} at^{2}

Where S is distance

u is the initial velocity

a is acceleration

and t is time

Since the players collide at the same time, then time spent by player 1 before collision equals time spent by player 2 before collision

That is, t₁ = t₂ = t

Where t₁ is the time spent by player 1

and t₂ is the time spent by player 2

For player 1

S = S₁

u = 0 m/s ( The player starts at rest)

a = 2.2 m/s²

Then,

S₁ = 0(t) + \frac{1}{2} (2.2)t^{2}

S₁ = 1.1t^{2}

t^{2} = \frac{S_{1} }{1.1}

For player 2

S = S₂

u = 0 m/s ( The player starts at rest)

a = 1.7 m/s²

Then,

S₂ = 0(t) + \frac{1}{2} (1.7)t^{2}

S₂ = 0.85t^{2}

t^{2} = \frac{S_{2} }{0.85}

Since the time spent by both players is equal, We can  write that

t^{2} = t^{2}

Then,

\frac{S_{1} }{1.1} =  \frac{S_{2} }{0.85}

0.85S_{1} = 1.1S_{2}

From, S₁ + S₂ = 41 m

S₂ = 41  - S₁

Then,

0.85S_{1} = 1.1 ( 41 -S_{1} )

0.85S_{1} = 45.1 - 1.1S_{1}

0.85S_{1} + 1.1S_{1} = 45.1 \\1.95S_{1} = 45.1\\S_{1} = \frac{45.1}{1.95} \\S_{1}  = 23.13 m

This is the distance covered by the first player. We can then put this value into t^{2} = \frac{S_{1} }{1.1} to determine how much time elapses before the players collide.

t^{2} = \frac{S_{1} }{1.1}

t^{2} = \frac{23.13 }{1.1}

t^{2} = 21.03\\t = \sqrt{21.03} \\t = 4.59 secs

Hence, the time that elapses before the players collide is 4.59 secs

3 0
4 years ago
Can someone rate my writing?<br><br>​
mylen [45]

Answer:

no writting shown but

Step-by-step explanation:

i guess its super cool encouragement is the most

3 0
3 years ago
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6 _[2/5 +{3 1/4 _(3/8+1/2)}] Solve it
Volgvan

6_2/5 + (3(1/4)_3/8 + 1/2

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3 years ago
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At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
SIZIF [17.4K]

Answer:

66?lol

Step-by-step explanation:

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3 years ago
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Please solve this for me thank you!
melamori03 [73]

Answer:

620.14

Step-by-step explanation:

Original Equation:

\frac{4y}{1.025^4}+y-2y(1.05)^2=1500

Calculate exponents

\frac{4y}{1.103812890625}+y-2y(1.1025)=1500\\

Simplify:

3.6238025791990193084270042653997y + y - 2.205y = 1500

Add like terms

2.4188025791990193084270042653997y \approx 1500

Divide both sides by 7.012

y\approx 213.91y\approx620.14

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2 years ago
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