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AlladinOne [14]
3 years ago
13

the three sides of a triangle are n, 3n+2, and 4n−4. If the perimeter of the triangle is 54cm, what is the length of each side?

Separate multiple entries with a comma.
Mathematics
2 answers:
taurus [48]3 years ago
8 0

The lengths of the sides are 7, 23 and 24.

In order to find this, we need to add all of the side lengths together and set equa to 54. This will allow us to solve for n.

n + 3n + 2 + 4n - 4 = 52

8n - 2 = 52

8n = 54

n = 7

This gives us the length of the first side. To solve for the others, plug 7 into the equations.

3n + 2

3(7) + 2

21 + 2

23

Then the next one.

4n - 4

4(7) - 4

28 - 4

24

Rom4ik [11]3 years ago
7 0

Answer:  7cm , 23 cm and 24.

Step-by-step explanation:

Given : The three sides of a triangle are n, 3n+2, and 4n−4.

We know that the perimeter of a polygon is the sum of all its sides.

Now, Perimeter of Triangle will be :-

P=n+3n+2+4n-4=8n-2

If the perimeter of the triangle is 54 cm , then we have

8n-2=54

Add 2 both sides, we get

8n=56

Divide both sides by 8, we get

n=7

Then , 3n+2 = 3(7)+2=23

4n-4 = 4(7)-4=24

Hence, the length of all its side will be 7cm , 23 cm and 24.

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Answer:

(a) This function is neither one-to-one nor onto.

(b) This function is neither one-to-one nor onto.

(c) This relation is not a function.

(d) The function is onto but not one-to-one.

Step-by-step explanation:

Given information: A = {1, 2, 3, 4, 5} and B = {a, b, c, d}

A relation is a function if and only if there exist a unique output for each input.

One-to-one : A function is one-to-one if every element of the function's codomain is the image of at most one element of its domain.

Onto : A function is onto if for every element y in the codomain Y of f there is at least one element x in the domain X of f such that f(x) = y.

(a)

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This relation is a function because all x-value has unique y-value.

The above function it not one-to-one because for more than one input we have same output (c have four domains).

The above function it not onto because all element of B are not have preimage (a and b have no preimage).

This function is neither one-to-one nor onto.

Similarly,

(b)

{(1, a ) ,(2, d ) ,(3, a ) ,(4, c ) }

This relation is a function because all x-value has unique y-value.

Here, a have more than one preimage and b have no preimage.

The function is neither one-to-one nor onto.

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{(1, d ) ,(2, d ) ,(3, a ) ,(4, b ) ,(4, d ) ,(4, c )} .

For x=4 we have for than one value of y.

Therefore this relation is not a function.

(d)

{(1, c ) ,(2, b ) ,(3, a ) ,(4, d ) ,(5, a ) }

This relation is a function because all x-value has unique y-value.

Here, a have two preimage. So, this function is not one-to-one.

All elements of B have preimage. So, this function is onto.

The function is onto but not one-to-one.

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