Question

Consider the equation an^2=182 where a is any number between 2 and 5 and n is a positive integer. What are the possible values of n???

Answer 1

I only found 9

Answer 2

an^2 = 182
Divide both sides by a
n^2 = 182/a
n = \(\sqrt{182/a}\)

cont..

Answer 3

an^2=182
n=√(182/a)

so you must find the values of a such that n stays between 2 and 5
for a=8 n=4.77 accepted
for a=45 n=2.01 accepted
for (a) less than 8 n>5
for (a) greater than 45 n<2

Answer 4

and for each value of a between 8 and 45 you have a different value of n between 2 and 5

Answer 5

great

Answer 6

do you need all the possible values or just the number of possible values??

Answer 7

a is any number between 2 & 5
182/5=36.4
182/2=91

Since √182/a is an integer 182/a is a perfect square (between 36.4 & 91)
there are 3 perfect square 49 ,64 ,81 between 36.4 and 91
therefore the values of n are \[\sqrt{49}=7 \ \sqrt{64}=8 \ \sqrt{81}=9\]

Answer 8

@hoblos all the possible values.

Answer 9

Diya, you still have the -ve values to include~ like -7x-7 =49

Answer 10

n is a (positive) integer @Callisto

Answer 11

Sorry... I was blind :(

Answer 12

No sorry =D

Answer 13

so what (n)*49=182????

Answer 14

its a*49=182

a is any number b/w 2&5

Answer 15

yes sorry, what a??

Answer 16

You can check the answer

a*49=182
a=182/49= 3.7 (approx.)

Answer 17

3.7 is a number between 2 & 5

Answer 18

Fantastic.

Answer 19

=)