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KatRina [158]
3 years ago
5

Lea walked 1/2 mile in 1/3 hour. How long will it take lea to walk 2 1/4 miles ?

Mathematics
2 answers:
cluponka [151]3 years ago
5 0

takes her 20 minutes to walk 1/2 a mile  

2 miles=80 minutes  

1/4= 10 minutes i think  

so 80+10=1 1/2

{This is not my work!!!}

aev [14]3 years ago
5 0

Step-by-step explanation:

Velocity = distance/time

1/3 h=20m

V=0.5/20=0.025m/m

So time= distance/velocity

T=2.25/0.25=90m

90m= 1.5h

I did it also by logic by using these steps:

1/2*(1/2 mile in 1/3 hour) = 1/4 mile in 1/6 hour

we have 9 quarter in 2 1/4

So 1/6 * 9 =1.5h

hope this helps !

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We know , the ratio of height and shadow , will be same for every object .

Let , length of shadow of pole is x .

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\dfrac{79.9}{136}=\dfrac{x}{40}\\\\x=\dfrac{79.9\times 40}{136}\ foot\\\\x=23.5\ foot

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Hence , this is the required solution .

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(a) Let R = {(a,b): a² + 3b &lt;= 12, a, b € z+} be a relation defined on z+)
grin007 [14]

Answer:

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Step-by-step explanation:

The relation R is an equivalence if it is reflexive, symmetric and transitive.

The order to options required to show that R is an equivalence relation are;

((a, b), (a, b)) ∈ R since a·b = b·a

Therefore, R is reflexive

If ((a, b), (c, d)) ∈ R then a·d = b·c, which gives c·b = d·a, then ((c, d), (a, b)) ∈ R

Therefore, R is symmetric

If ((c, d), (e, f)) ∈ R, and ((a, b), (c, d)) ∈ R therefore, c·f = d·e, and a·d = b·c

Multiplying gives, a·f·c·d = b·e·c·d, which gives, a·f = b·e, then ((a, b), (e, f)) ∈R

Therefore R is transitive

From the above proofs, the relation R is reflexive, symmetric, and transitive, therefore, R is an equivalent relation.

Reasons:

Prove that the relation R is reflexive

Reflexive property is a property is the property that a number has a value that it posses (it is equal to itself)

The given relation is ((a, b), (c, d)) ∈ R if and only if a·d = b·c

By multiplication property of equality; a·b = b·a

Therefore;

((a, b), (a, b)) ∈ R

The relation, R, is reflexive.

Prove that the relation, R, is symmetric

Given that if ((a, b), (c, d)) ∈ R then we have, a·d = b·c

Therefore, c·b = d·a implies ((c, d), (a, b)) ∈ R

((a, b), (c, d)) and ((c, d), (a, b)) are symmetric.

Therefore, the relation, R, is symmetric.

Prove that R is transitive

Symbolically, transitive property is as follows; If x = y, and y = z, then x = z

From the given relation, ((a, b), (c, d)) ∈ R, then a·d = b·c

Therefore, ((c, d), (e, f)) ∈ R, then c·f = d·e

By multiplication, a·d × c·f = b·c × d·e

a·d·c·f = b·c·d·e

Therefore;

a·f·c·d = b·e·c·d

a·f = b·e

Which gives;

((a, b), (e, f)) ∈ R, therefore, the relation, R, is transitive.

Therefore;

R is an equivalence relation, since R is reflexive, symmetric, and transitive.

Based on a similar question posted online, it is required to rank the given options in the order to show that R is an equivalence relation.

Learn more about equivalent relations here:

brainly.com/question/1503196

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