You are presented with 9 coins from an archeological dig near the highlands of Tibet. One of them contains a precious jewel that can make you extremely wealthy. The coin with the jewel is slightly heavier than the others, the other eight are identical. You can only choose once so you are given 2 scales so that you can weigh the coins. Here’s the dilemma, each balance scale can only be weighed once. Is it possible to select the slightly heavier coin containing the jewel using each of the two scales only once? Explain why or why not

just get this app called photo math it should be able to do those

step-by-step explanation:

one day the were 5 ducks, and their king told them to multiply their kingdom. so each went their own seperate way and met 4 other ducks on their journey.

when the ducks returned to their kingdom, it was discovered that there were 5 groups of 5 ducks. when the kind counted the number of ducks in total (including the ones he sent out earlier) he counted 25.

therefore 5 times 5 is 25.

the end. : )

asnwer is (g o h o f)(x) = 2x - 6/x + 7

step-by-step explanation:

given,

f(x) = x - 3/x

g(x) = x + 3

h(x) = 2x + 1

we need to find : (g o h o f)(x)

we can write (g o h o f)(x) = g[ h{ f(x) } ]

so first we find h{f(x)}

h{f(x)} = f(x) + 3 = (x - 3/x) + 3

h{f(x)} = x - 3/x + 3

now we find g[ h{ f(x) } ].

g[ h{ f(x) } ] = g(x - 3/x + 3)       {plug h{f(x)} = x - 3/x + 3 in g[h{f(x)}] }

g(x - 3/x + 3) = 2 (x - 3/x + 3) + 1

                    = 2x - 6/x + 6 + 1

                    = 2x - 6/x + 7

hence we get (g o h o f)(x) = 2x - 6/x + 7