All categories

3 years ago

Which set of ordered pairs represents a function, and how does it represent that function?

1 answer:

5 0

B is the only one thats a function because it has one X for every Y while the other options have multiple X's for one Y. That does not pass the vertical line test and therfore not a function.

You might be interested in

Four friends ordered a pizza cut into eight slices. The whole pizza was eaten, with each friend eating a different number of sli

raketka [301]

Answer:

{0,1,3,4}

Step-by-step explanation:

Given, total no of slices=8.

Total no of friends=4.

also each friend didnt ate more than the other three.

let a,b,c,d be no of slices ate by the friends 1,2,3,4 respectively.

given sum of a+b+c+d=8.

now applying the second condition

$a\leq b+c+d \\b\leq c+d+a\\c\leq a+b+d\\d\leq a+b+c$

so each of a,b,c,d must be less than or equal to 4.

which means 5 values are possible{0,1,2,3,4}.from this set you have to select 4 values such that their sum is 8.hence, those values are {0,1,3,4}.

7 0

3 years ago

Read 2 more answers

When Devon cashed a $450 check at the bank, the teller gave him 18 bills, all $20 bills and $50 bills. Which system of equations

Neporo4naja [7]

B is the answer to the equation

7 0

3 years ago

There are fifteen teams in a high school baseball league. How many different orders of finish are possible for the first four p

ankoles [38]

There are 32,760 different orders for the first four positions.

This is a permutation where repetition is not allowed; the formula for that is:

$_nP_r=\frac{n!}{(n-r)!}$

For our problem, we have:
$_{15}P_4=\frac{15!}{(15-4)!}=\frac{15!}{11!}=32760$

3 0

3 years ago

Product of two natural numbers p and q is 590 and their hcf is 59. how many set of values of p and q is possible

djverab [1.8K]

$\text{Let the product of two natural numbers p and q is 590, and their HCF is 59}\\
\\
\text{we know that the product of LCM and HCF of any two numbers is equal}\\
\text{to the product of the numbers. that is}\\
\\
\text{HCF}\times \text{ LCM}=p\times q\\
\\
\Rightarrow 59 \times \text{LCM}=590\\
\\
\Rightarrow \text{LCM}=\frac{590}{59}\\
\\
\Rightarrow \text{LCM}=10\\
\\
\text{for any two natural numbers, their Least Common Multiple (LCM) is always}$

$\text{greater than their HCF.}\\
\\
\text{but here we can see that }LCM$

Hence there is no such natural numbers exist.

7 0

3 years ago

Help Me Please! Will reward Branliest!

alexgriva [62]

Answer:

Hey the person from the guy or gurl who told me to join the zoom. I wasn't rlly gonna join. I think its thrid option

8 0

3 years ago