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inna [77]
2 years ago
9

I dont know the answer to this question do you 3 x blank = 1

Mathematics
2 answers:
faltersainse [42]2 years ago
3 0

Answer:

y = \frac13

Step-by-step explanation:

let the blank be y

3 * y = 1

3y = 1

Now, divide both sides by 3:

3y ÷ 3 = 1 ÷ 3

y = 1/3

-Chetan K

Anon25 [30]2 years ago
3 0

Answer:

<u>Value of the blank </u>(We assumed it as x while solving)<u> :-</u>

\boxed {\tt x =  \cfrac{1}{3}}

<u>Actual answer :-</u>

\boxed{\tt \: 3 \times  \boxed{\cfrac{1}{3}}  = 1}

Step-by-step explanation:

<u>Given equation :-</u>

\sf 3 \times ? = 1

(<u>Note</u>: I wrote "?" instead of "blank".)

<em>We need to find the value of the blank on this equation</em>.

<u>Solution:-</u>

To solve Further let's us assume that the value of the blank be x.

So, blank = x

The equation will be . . .

\sf \longmapsto3 \times x = 1

Let's solve out this equation !

\sf \implies3 \times x = 1

<u>Multiply 3 and x on the LHS :-</u>

\sf \implies3 x = 1

<em>Hence the new equation would be </em>3x=1.

<u>Now, Divide both sides by 3 :-</u>

\sf \implies \:  \cfrac{3x}{3}  =  \cfrac{1}{3}

<u>Simplify this equation :-</u>

Cancel 3 and 3 on the LHS, Leave x . 1/3 can't be cancelled.

\sf \implies \:  \cfrac{ \cancel3}{  \cancel3x}  = \cfrac{1}{3}

\sf \implies1x =  \cfrac{1}{3}

As we know that x = 1x So,

\sf \implies \: x =  \cfrac{1}{3}

<em>This equation can't be simplified more.</em>

Hence, the value of x would be 1/3.

<u>Now,Put the value of </u><u>x</u><u> </u><u>(1/3) on the given equation:-</u>

\sf \implies \:  3 \times  \: \cfrac{1}{3}  \:   = 1

We are done !

______________________________________

I hope this helps !

Let me know if you have any questions.

\Huge\sf :)

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Len [333]

Answer:

The relation is not a function

The domain is {1, 2, 3}

The range is {3, 4, 5}

Step-by-step explanation:

A relation of a set of ordered pairs x and y is a function if

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  • x appears once in ordered pairs

<u><em>Examples:</em></u>

  • The relation {(1, 2), (-2, 3), (4, 5)} is a function because every x has only one value of y (x = 1 has y = 2, x = -2 has y = 3, x = 4 has y = 5)
  • The relation {(1, 2), (-2, 3), (1, 5)} is not a function because one x has two values of y (x = 1 has values of y = 2 and 5)
  • The domain is the set of values of x
  • The range is the set of values of y

Let us solve the question

∵ The relation = {(1, 3), (2, 3), (3, 4), (2, 5)}

∵ x = 1 has y = 3

∵ x = 2 has y = 3

∵ x = 3 has y = 4

∵ x = 2 has y = 5

→ One x appears twice in the ordered pairs

∵ x = 2 has y = 3 and 5

∴ The relation is not a function because one x has two values of y

∵ The domain is the set of values of x

∴ The domain = {1, 2, 3}

∵ The range is the set of values of y

∴ The range = {3, 4, 5}

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Answer:

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Step-by-step explanation:

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