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Bad White [126]
3 years ago
12

Find the unit rate for 50 miles on 2 1/2 gallons

Mathematics
1 answer:
Anarel [89]3 years ago
6 0
Unit rate is miles per 1 gallon
50mi/2 and 1/2 gallon
50mi/2.5 gallon
times 10/10
500mi/25 gallon
20mi/1gallon

answer is 20mi/gal
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on our way to visit her parents Jennifer Drive 265 miles in 5 hours what is the average rate of speed in miles per hou
Setler79 [48]
The average speed would be 53 miles per hour, hope this helped, if you need me to explain then here is the way, so all you do is divide 265 by 5 and what I did to help make it faster was split 265 into 250 and 15 so 250 divided by 5 is 50 and 15 divided by 5 is 3 so 50 + 3 = 53!
6 0
4 years ago
Charlene bought a used.It cost $43.00.A new skateboard costs 3 times as much as the used one Charlene bought.How much does a new
vovangra [49]
Multiply it by 3.
43.00*3=129=priceNewSkateboard
6 0
3 years ago
Help me please explain do not need work just explain THANK YOU I NEED HELP
Mnenie [13.5K]
Alright, 

you will need to solve for each variable to find if they got the equation correctly or not.. 

Let's start with h 
isolate h 
(1/2)h = A/(b1+b2)
Now we need to get rid of 1/2 we may do so by multiplying both sides by 2 

h = 2(A/(b1+b2))

And that's not how they did it for h so Option D doesn't apply 


Let's see C
Also doesn't apply since they multiplied 2 only by A rather than A/h

Let's see B
Also not correct they made the same mistake as with option C

Let's see A
again same mistake 
when multiplying both sides by 2 
the 2 should be as following 
2\frac{A}{h}

I believe you should contact your instructor and explain that non of the options is right. 

6 0
3 years ago
Please help me thanks!​
lions [1.4K]
5 x 16 x 4 = 320 m^3
3 0
3 years ago
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)\}

5 0
3 years ago
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