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hichkok12 [17]
3 years ago
11

Find the surface area of the right triangular prism shown below:

Mathematics
2 answers:
Elza [17]3 years ago
4 0

Answer:

33

Step-by-step explanation:

Because 5 * 5/2 is 12.5

5 * 3/2 is 7.5

5 * 2 is 10

2 * 3/2 is 3 and instead of dividing by 2 you don't cause there is 2 triangles

Genrish500 [490]3 years ago
4 0
The answer is 33, as shown above
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.2x+.04x=4.8 solve for x…………………
vaieri [72.5K]
X=20
Combine like terms and divide both sides by .24
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3 years ago
Read 2 more answers
Helppp !!!
Dmitry_Shevchenko [17]

Answer: 2 3/4

Step-by-step explanation: 1/4 + 3/4 is 1 whole plus 1/4+1/4 would equal the 1/2 you need.

So we are up to 3/4 already. 1 3/4 + 3/4 = 2 1/2

We need to add 2 more to get to 4 1/2

So the answer would be 2 3/4

2 3/4 + 1 3/4 = 4 1/2

4 0
3 years ago
Help help help help <br> Answer please
Ludmilka [50]

Answer:

Mark talked on the phone for 120 minutes.

Step-by-step explanation:

18 - 12 = 6

6 ÷ .05 = <u>1</u><u>2</u><u>0</u>

6 0
3 years ago
Well happy New Year everyone. I've been thinking about the upcoming year "2021" and found something interesting. Reduced to its
Anarel [89]

Answer:

As 2021 and 1927 does not have any product of primes

HCF and LCM of 2020 and 1926 is shown below

Step-by-step explanation:

2020 = 2^2 × 5 × 101

1926 = 2 x 3^2 x 107

2  is the HCF

2 x 2 x 3 x 3 x 5 x 101 x 107  = 1945260

1945260 is the LCM

2022 and 1928 both have product of primes and their HCF and LCM is

1949208 is the LCF and 2 is the HCF

1949208 - 1945260 = 3948

3948/2 = 1974

2022-1928 = 94

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Happy New Year to you.

5 0
3 years ago
Identify at least one Hamilton path and at least one Hamilton circuit
Anna71 [15]

We will investigate how to determine Hamilton paths and circuits

Hamilton path: A path that connect each vertex/point once without repetition of a point/vertex. However, the starting and ending point/vertex can be different.

Hamilton circuit: A path that connect each vertex/point once without repetition of a point/vertex. However, the starting and ending point/vertex must be the same!

As the starting point we can choose any of the points. We will choose point ( F ) and trace a path as follows:

F\to D\to E\to C\to A\to B\to F

The above path covers all the vertices/points with the starting and ending point/vertex to be ( F ). Such a path is called a Hamilton circuit per definition.

We will choose a different point now. Lets choose ( E ) as our starting point and trace the path as follows:

E\to D\to F\to B\to A->C

The above path covers all the vertices/points with the starting and ending point/vertex are different with be ( E ) and ( C ), respectively. Such a path is called a Hamilton path per definition.

One more thing to note is that all Hamilton circuits can be converted into a Hamilton path like follows:

F\to D\to E\to C\to A\to B

The above path is a hamilton path that can be formed from the Hamilton circuit example.

But its not necessary for all Hamilton paths to form a Hamilton circuit! Unfortunately, this is not the case in the network given. Every point is in a closed loop i.e there is no loose end/vertex that is not connected by any other vertex.

4 0
1 year ago
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