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gregori [183]
3 years ago
15

The chart below lists the original and sale prices of items at a clothing store.

Mathematics
2 answers:
Anon25 [30]3 years ago
7 0

Answer:

C STAY SAFE!!!

Step-by-step explanation:

Ok we know this cant be A the reason is It says tha the original price is increasing so thats FALSE... its trying to trick you so no

The second choice says The sale price is always less than the original price. well take a look at the sale prices are they? Obiously not so False

Ok the third option For every original price, there is exactly one sale price. well this is true ask yourself each it helps.

Last option The sales price is never less than zero. erm   FALSE OBIOUSLY THIS IS TRUE JUST NOT TRUE ITS WRONG        

THE ANSER IS C

Tom [10]3 years ago
5 0

Answer: C) For every original price, there is exactly one sale price.

For any function, we always have any input go to exactly one output. The original price is the input while the output is the sale price. If we had an original price of say $100, and two sale prices of $90 and $80, then the question would be "which is the true sale price?" and it would be ambiguous. This is one example of how useful it is to have one output for any input. The input in question must be in the domain.

As the table shows, we do not have any repeated original prices leading to different sale prices.

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Effectus [21]

Answer:

12 and 2. or. 4 and 3

Step-by-step explanation:

the LCM of 12 and 2 is 12

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Answers because I need help
kramer

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Step-by-step explanation:

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Read 2 more answers
Use induction to prove that 2? ?? for any integer n>0 . Indicate type of induction used.
Hoochie [10]

Answer with explanation:

The given statement is which we have to prove by the principal of Mathematical Induction

    2^{n}>n

1.→For, n=1

L H S =2

R H S=1

2>1

L H S> R H S

So,the Statement is true for , n=1.

2.⇒Let the statement is true for, n=k.

      2^{k}>k

                   ---------------------------------------(1)

3⇒Now, we will prove that the mathematical statement  is true for, n=k+1.

     \rightarrow 2^{k+1}>k+1\\\\L H S=\rightarrow 2^{k+1}=2^{k}\times 2\\\\\text{Using 1}\\\\2^{k}>k\\\\\text{Multiplying both sides by 2}\\\\2^{k+1}>2k\\\\As, 2 k=k+k,\text{Which will be always greater than }k+1.\\\\\rightarrow 2 k>k+1\\\\\rightarrow2^{k+1}>k+1

Hence it is true for, n=k+1.

So,we have proved the statement with the help of mathematical Induction, which is

      2^{k}>k

                 

   

3 0
3 years ago
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