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2 years ago

Dimensional analysis: 1.35 kilometers per second= ? miles per hour

Mathematics

2 answers:

kramer2 years ago

5 0

Answer:

50.2 miles/h.

Step-by-step explanation:

Turn km to miles:

1 km = 0.621 miles

1.35 km = ?

1.35 × 0.621 = 0.838 miles/s

0.838 miles = 1 second

? = 1 hour

1 hour = 60 minutes

60 × 0.838

= 50.28 miles/h

Hope this is helpful.

Elza [17]2 years ago

4 0

Answer:

Hey there!

1.35 km/s would equal 3020 miles per hour.

Let me know if this helps :)

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Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:

professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is $\{1,2,3,4\}$ .

a) We need to find a relation R reflexive and transitive that contain the relation $R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}$

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

1R4 and 4R1, then 1 must be related with itself.
4R1 and 1R4, then 4 must be related with itself.
4R1 and 1R2, then 4 must be related with 2.

Therefore $\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\}$ is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

since 1R2, then 2 must be related with 1.
since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

1R2 and 2R1, then 1 must be related with itself.
4R1 and 1R4, then 4 must be related with itself.
2R1 and 1R4, then 2 must be related with 4.
4R1 and 1R2, then 4 must be related with 2.
2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

$\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}$

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

1 must be related with 1,

2 must be related with 2,

3 must be related with 3,

4 must be related with 4

For be symmetric

since 1R2, 2 must be related with 1,
since 1R4, 4 must be related with 1.

For be transitive

Since 4R1 and 1R2, 4 must be related with 2,
since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

$\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}$

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elena55 [62]

Answer:

Option A,$320.98 is correct

Step-by-step explanation:

The new balance on Elijah's credit is the sum of the last balance plus new charges minus payments made.

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new charges were $213.40

payment made by Elijah was $45

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The rationale for this is that the new charges on the card account add to previous balance owed by Elijah while the payment of $45 made reduces overall balance.

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erma4kov [3.2K]

Equation:

a = 5120

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f (n ) = 5120 * 1 . 2^n - 1

N = 4 years

f ( 4 ) 5120 * 1 . 2^4 - 1

5120 * 1 . 2^3

We want to know, The number of Highway Accidents in Four Years.

Answer: 8,847 Highway Accidents total:

Hope that helps!!!!!!!!!!! : )

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Answer:

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Step-by-step explanation:

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