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Harman [31]
2 years ago
12

Find the surface area of the composite solid and round to the nearest tenth. Explain your answer.

Mathematics
1 answer:
Orlov [11]2 years ago
7 0
Well, notice the composite is really just 4 triangles atop sitting on top of 4 rectangles, and all of them area stacked up at the edges.

so, for the rectangle's sides,

front and back are two 6x3 rectangles

left and right are two 6x3 rectangles

the bottom part is a 6x6 rectangle

now, we don't include the 6x6 rectangle that's touching the triangles, because that's inside area, and is not SURFACE area, so we nevermind that one.

now, the triangles are just four triangles with a base of 6, and a height of 4, in red noted there.

so, just get the area of all those rectangles and the triangles, sum them up and that's the surface area of the composite,

\bf \stackrel{lateral~rectangles}{4(6\cdot 3)}~~+~~\stackrel{bottom~rectangle}{(6\cdot 6)}~~+~~\stackrel{triangles}{4\left( \cfrac{1}{2}(6)(4) \right)}
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A bag contains different colored candies. There are 50 candies in the bag, 28 are red, 10 are blue, 8 are green and 4 are yellow
Mrrafil [7]

Answer:

\displaystyle \frac{54}{5405}.

Step-by-step explanation:

How many unique combinations are possible in total?

This question takes 5 objects randomly out of a bag of 50 objects. The order in which these objects come out doesn't matter. Therefore, the number of unique choices possible will the sames as the combination

\displaystyle \left(50\atop 5\right) = 2,118,760.

How many out of that 2,118,760 combinations will satisfy the request?

Number of ways to choose 2 red candies out a batch of 28:

\displaystyle \left( 28\atop 2\right) = 378.

Number of ways to choose 3 green candies out of a batch of 8:

\displaystyle \left(8\atop 3\right)=56.

However, choosing two red candies out of a batch of 28 red candies does not influence the number of ways of choosing three green candies out of a batch of 8 green candies. The number of ways of choosing 2 red candies and 3 green candies will be the product of the two numbers of ways of choosing

\displaystyle \left( 28\atop 2\right) \cdot \left(8\atop 3\right) = 378\times 56 = 21,168.

The probability that the 5 candies chosen out of the 50 contain 2 red and 3 green will be:

\displaystyle \frac{21,168}{2,118,760} = \frac{54}{5405}.

3 0
3 years ago
What is half of pi/3?
Alborosie
Half of pi/3 means 1/2 times pi/3=pi/6
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