On Saturday's lunch rush at Ada's diner, they sold 295 sandwiches. Of those sandwiches sold: 68 sandwiches included turkey 53 sandwiches included cheese 15 sandwiches included both turkey and cheese 9 sandwiches included turkey and sprouts 7 sandwiches included both cheese and sprouts 5 sandwiches included all three ingredients 18 sandwiches had none of these ingredients (turkey, cheese, or sprouts) How many sandwiches only included sprouts (of these three ingredients)?

182 sandwiches

Step-by-step explanation:

How many sandwiches only included sprouts (of these three ingredients)

Total sandwiches = 295

68sandwiches included turkey

53 sandwiches included cheese

15 sandwiches included both turkey and cheese

9 sandwiches included turkey and sprouts

7 sandwiches included both cheese and sprouts

5 sandwiches included all three ingredients

18 sandwiches had none of these ingredients (turkey, cheese, or sprouts)

Let's represent

Turkey = T

Sprouts = S

Cheese = C

n (T ∪ S ∪ C) - n(none of the ingredients) = n(T ) + n ( S ) + n (C) - n ( T ∩ S) - n ( T ∩ C) - n ( C ∩ S) + n (T ∩ S ∩ C )

= 295 - 18 = 68 + 53+ n(S) - 9 - 15 - 7+ 5

= 277 = 121 + n(S) - 26

= 277 - 121 + 26= n(S)

= 303 - 121 = n(S)

= 182 = n(s)

Hence, only 182 sandwiches had only sprouts

i need a picure

step-by-step explanation:

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.