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pav-90 [236]
3 years ago
5

A bathtub is filled with 38 1/3 gallons of water. If 2 3/5 gallons splash out, how much water is left in the tube

Mathematics
1 answer:
musickatia [10]3 years ago
3 0

So let's convert these to improper fractions:

38\frac{1}{3}=\frac{115}{3}

2\frac{3}{5}=\frac{13}{5}

Then let's convert these fractions so that they have a denominator of 15 so we can better compare these numbers:

\frac{115}{3}=\frac{575}{15}

\frac{13}{5}=\frac{39}{15}

So when we have \frac{39}{15} gallons splash out, we are left with \frac{536}{15} gallons or 35\frac{11}{15} gallons in the tub.

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John bikes at a speed of 10 miles per hour. His friend Brian is slower, his biking speed is only 8 miles per hour. d Tomorrow bo
skad [1K]

Answer:

  • 6 miles

Step-by-step explanation:

Given

  • John's speed = 10 mph
  • Brian's speed = 8 mph
  • Time 3 hours

Solution

<u>The difference in distance covered:</u>

  • 3*(10 - 8) = 6 miles
3 0
3 years ago
. Solve for x.<br> x – 2 = 30
nadya68 [22]

Answer:

X=32

Step-by-step explanation:

32-2=30

6 0
2 years ago
Which of these relations on{0,1,2,3}are partial orderings? Determine the properties of a partial ordering that the others lack.
omeli [17]

Step-by-step explanation:

A = {0,1,2,3}

a): R = {(0,0),(2,2),(3,3)}

R is antisymmetric, because if the (a,b)∈R, than a=b.

R is not reflexive, because (1,1) ∉ R while 1 ∈ A.

R is transitive, because if the (a,b)∈R and (b, c) ∈ R, than a=b=c and (a,c)=(a,a)∈R.

R is not portable ordering because R is not reflexive.

b): R = {(0,0),(1,1),(2,0),(2,2),(2,3),(3,3)}

R is antisymmetric, because if the (a,b)∈R and if the (b, a) ∈ R, than a=b (since (2,0) ∈ R and (0,2) ∉ R; and (2,3) ∈ R and (3,2) ∉ R )

R is reflexive, because (a,a) ∈ R of every element a ∈ A.

R is transitive , because if the (a,b)∈R and if the ( b , c )∈R . then a = b or b = c ( since there are only two element not of the form ( a , a ) and that pair does not satisfy ( a,b ) ∈ R and ( b , a ) ∈ R ), which implies ( a , c ) = ( b , c ) ∈ R or ( a , c ) = ( a , b ) ∈ R.

R is a partial ordering, because R is reflexive, antisymmetric and transitive.

c): R =  {(0,0),(1,1),(1,2),(2,2),(3,1),(3,3)}

R is reflexive, because (a,a)∈R of every element a ∈ A.

R is antisymmetric, because if the ( a , b )∈R and if the ( b , a )∈R . then a = b ( since ( 1 , 2 )∈R and ( 2 , 1 ) ∉ R; ( 3 , 1 ) ∈ R and ( 1 , 3 ) ∉ R ).  

R is not transitive , because ( 3 , 1 ) ∈ R and ( 1 , 2 )∈R, while ( 3 , 2 ) ∈ R.

R is not a partial ordering. because R is not transitive .

d): R =  {(0,0),(1,1),(1,2),(1,3),(2,0),(2,2),(2,3), (3,0),(3,3)}

R is the reflexive, because ( a , a )∈R of every elements∈A.

R is the antisymmetric, because if the ( a , b )∈R and if the ( b , a )∈R, then a = b ( since ( 1 . 2 )∈R and ( 2 . 1 )∉R; similarly, all other elements not of the form (a,a) ).

R is not transitive, because ( 1 , 2 )∈R and ( 2 , 0 )∈R, while ( 1 . 0 )∉R.

R is not a partial ordering, because R is not transitive,

e):  R = { ( 0 , 0 ) , ( 0, 1 ) , ( 0 , 2 ) , ( 0 , 3 ) , ( 1 , 0 ) , ( 1 , 1 ) , ( 1 , 2 ) , ( 1 , 3 ) , ( 2 , 0 ) , ( 2 , 2 ) , ( 3 , 3 ) }

R is the reflexive , because ( a , a )∈R of every element a∈A .

R is not antisymmetric, because ( 1 , 0 )∈R and ( 0 , 1 )∈R while 0 is not equal to 1.

R is not transitive, because ( 2 , 0 )∈Rand ( 0 , 3 )∈R, while ( 2 , 3 )∉R .

R is not a partial ordering, because R is not the antisymmetric and not the transitive.

3 0
3 years ago
1) Your friend Taylor missed class today and needs some help identifying solutions of systems. Explain to Taylor where to find t
Mashcka [7]

Answer:

Ques 1:

Let us consider the system of equations,

2x+y=3 and 3x-y=2

It is known that 'The intersection points of the graphs of the system of equations are the solutions of the system of equations'.

So, after plotting the given graphs, we see that from the first graph,

The only intersection point is (1,1).

Thus, the solution of the system is x= 1 and y= 1.

Ques 2:

We have the functions, f(x)= 2x+1 and g(x)= 2x^2+1.

Plotting the graphs of both the functions is shown by graph 2, we see that,

The intersection points are (0,1) and (1,3)

Hence, the function have equal values only at x= 0 and x= 1 but not at x= 3.

8 0
3 years ago
ILL put you as brainliest
ICE Princess25 [194]
Just replace x with whatever , and see if it works. If it works then it is a solution; otherwise it's not.

1^2 is not 2
(1/2)^2 isn't 2 either...
and so on
3 0
3 years ago
Read 2 more answers
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