There are 12 m&m.
Because there are 6 skittles to every 4 m&m
There are 18 skittles.
So there must be 12 skittles
∆PQR and ∆SQT are similar, so corresponding sides occur in fixed ratios with one another. This means
SQ / PQ = ST / PR
Solve for x :
8 / ((x + 5) + 8) = (x - 9) / 21
8 / (x + 13) = (x - 9) / 21
168 = (x - 9) (x + 13)
168 = x² + 4x - 117
x² + 4x - 285 = 0
(x + 19) (x - 15) = 0
⇒ x = -19 or x = 15
But x can't be smaller than 9, otherwise ST = x - 9 would be a negative length. So x = 15.
The answer to your question is 3
Answer: Box3 has 120 oranges.
Box1 has 150 oranges.
Box has 130 oranges.
Step-by-step explanation:
I will answer this in English.
The question says:
3 boxes have 400 oranges.
The first one has 20 more than the second and 30 more than the third.
So we have 3 equations:
box1 + box2 + box3 = 400
box1 = box2 + 20
box1 = box3 + 30
Now, we can take the variable box1 in the third equation and replace it on the other two.
box3 + 30 + box2 + box3 = 400
box3 + 30 = box2 + 20
Now we can isolate one of the variables in the second equation, let's isolate box2.
box2 = box3 + 30 - 20 = box3 + 10
now we can replace it in the other equation:
box3 + 30 + box2 + box3 = 400
box3 + 30 + box3 + 10 + box3 = 400
3*box3 + 40 = 400
3*box3 = 400 - 40 = 360
box3 = 360/3 = 120.
Box3 has 120 oranges.
Box1 has 120 + 30 = 150 oranges.
Box has 150 - 20 = 130 oranges.
