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saul85 [17]
2 years ago
12

We can calculate the depth ddd of snow, in centimeters, that accumulates in Harper's yard during the first hhh hours of a snowst

orm using the equation d=5hd=5hd, equals, 5, h. How many hours does it take for 111 centimeter of snow to accumulate in Harper's yard? hours How many centimeters of snow accumulate per hour? centimeters
Mathematics
2 answers:
Scorpion4ik [409]2 years ago
6 0

Answer:

a) Number of hours it takes 1 centimetre of snow to form in Harper's yard = (1/5) hour = 0.20 hour

b) Centimetres of snow that accumulate per hour = 5 cm

Step-by-step explanation:

Complete Question

We can calculate the depth d of snow, in centimeters, that accumulates in Harper's yard during the first h hours of a snowstorm using the equation d=5h.

a) How many hours does it take for 1 centimeter of snow to accumulate in Harper's yard? hours

b) How many centimeters of snow accumulate per hour? centimeters

Solution

The depth of snow, d, in centimetres that accumulates in Harper's yard in h hours is given d = 5h

a) Number of hours it takes 1 centimetre of snow to form in Harper's yard.

d = 5h

d = 1 cm

h = ?

1 = 5h

h = (1/5) = 0.20 hour

b) Centimetres of snow that accumulate per hour.

d = 5h

In 1 hour, h = 1 hour

d = ?

d = 5 × 1 = 5 cm

Hope this Helps!!!

vodomira [7]2 years ago
3 0

Answer:

1/5 hours

5 centimeters

Step-by-step explanation:

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Answer:

g(2) = 3

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Step-by-step explanation:

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Given that a function, g, has a domain of -1 ≤ x ≤ 4 and a range of 0 ≤ g(x) ≤ 18 and that g(-1) = 2 and g(2) = 8, select the st
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Options

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Answer:

(D)g(3) = 18

Step-by-step explanation:

Given that the function, g, has a domain of -1 ≤ x ≤ 4 and a range of                       0 ≤ g(x) ≤ 18 and that g(-1) = 2 and g(2) = 8

Then the following properties must hold

  1. The value(s) of x must be between -1 and 4
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We consider the options and state why they are true or otherwise.

<u>Option A: g(5)=12</u>

The value of x=5. This contradicts property 1 stated above. Therefore, it is not true.

<u>Option B: g(1) = -2 </u>

The value of g(x)=-2. This contradicts property 2 stated above. Therefore, it is not true.

<u>Option C: g(2) = 4 </u>

The value of g(2)=4. However by property 4 stated above, g(2)=9. Therefore, it is not true.

<u>Option D: g(3) = 18</u>

This statement can be true as its domain is in between -1 and 4 and its range is in between 0 and 18.

Therefore, Option D could be true.

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Step-by-step explanation:

Given: Lenin is preparing dinner plates. He has 12 pieces of chicken and 16 rolls.

To make all the plates identical without any food left over, the greatest number of plates Lenin can prepare = G.C.D.(12,16)=4

The number of pieces of chicken in each plate = \dfrac{12}{4}=3

The number of pieces of rolls in each plate = \dfrac{16}{4}=4

So,  the greatest number of plates Lenin can prepare = 4

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Which of these relations on{0,1,2,3}are partial orderings? Determine the properties of a partial ordering that the others lack.
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Step-by-step explanation:

A = {0,1,2,3}

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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.

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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 ).  

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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.

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Answer:

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506/723=0.69

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