Mrs cohan is making halloween bags for her party. she has 70 milky way bars and 105 pieces of bubble gum. how many pieces of milky way bars and bubble gum can be in each bag.

5 Halloween bags containing 14 bars and 21 gum pieces, 7 Halloween bags containing 10 bars and 15 gum pieces and 35 bags containing 2 bars and 3 gum pieces can be created using 70 Milky Way bars and 105 pieces of bubble gum.

Solution:

Mrs Cohan is making Halloween bags for her party.

Number of Milky Way bars with Mrs Cohan = 70

Number of bubble gum pieces with Mrs Cohan = 105

We need to distribute 70 bars and 105 bubble gum pieces in each bag

On factorizing 70 we get

On factorizing 105 we get

The common factors of 70 and 105 are 5, 7 and 35  

So 70 bars and 105 bubble gum pieces can be equally distributed among 5 bags, 7 bags and 35 bags.

⇒ Number of milky bar in each bag when 5 bags are created

⇒ Number of bubble gum pieces in each bag when 5 bags are created

⇒ Number of milky bar in each bag when 7 bags are created

⇒ Number of bubble gum pieces in each bag when 5 bags are created

⇒ Number of milky bar in each bag when 35 bags are created

⇒ Number of bubble gum pieces in each bag when 35 bags are created

Hence ⇒ 5 Halloween bags containing 14 bars and 21 gum pieces

⇒ 7 Halloween bags containing 10 bars and 15 gum pieces

⇒ 35 bags containing 2 bars and 3 gum pieces can be created using 70 Milky Way bars and 105 pieces of bubble gum.

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!