The answer would be C. 3/12 (Reduced 1/4) (Decimal 0.25) I hope this helped ^^
Part A: each tricycle has three wheels, so with 48 wheels the number of tricycles was a =48/3=16 tricycles.
t=w/3 (the number of tricycles is the number of wheels divided by 3)
Part B:
The number of seats:
24=b+a (so b=24-a)
The number of seats is the sum of one seat per bicycle and one seat per a tricycle
also, 61=2a+3b (the number of wheels)
So we have:
24=b+a
b=24-a
We can substitute this for b:
61=2a+3(24-a)
and solve:
61=2a+3*24-3a
61=72-a
a=72-61
a=11
There were 11 bicycles!!
and there were 24-11 tricycles, so 13 tricycles.
Part C: each of the bikes has only one front-steering handlebar, so there were a total of 144 vehicles:
a+b+c=144
There were 378 pedals. And the number of pedals is:
2a+2b+4c=378 (the numbers 2,2,4 represent the number of pedals per vehicle)
divide by 2:
a+b+2c=189
Now, we have
a+b+2c=189
and
a+b+c=144
and we can subtract them from each other:
a+b+c-(a+b+2c)=144-189
-c=45
c=45, so there were 45 tandem bicycles!
(this also means that a+b=144-45, that is a+b=99)
now the wheels:
3a+2b+2c=320
Let's substitute c:
3a+2b+90=320
which is
3a+2b=240
We also know that a+b=99, so we can substract this from this equation:
3a+2b+-a-b=240-99
2a+b=141
and again:
2a+b-a-b=141-99
a=42 - there were 42 trycicles!!!
And the bicycles were the rest:
99-42=57 bycicles
Answer: ∠OZP = 62 ∠PZQ = 63
<u>Step-by-step explanation:</u>
Use the Angle Addition Postulate
∠OZP + ∠PZQ = ∠OZQ
(4r + 2) + (5r - 12) = 125
9r - 10 = 125
9r = 135
r = 15
∠OZP = 4r + 2
= 4(15) + 2
= 60 + 2
= 62
∠PZQ = 5r - 12
= 5(15) - 12
= 75 - 12
= 63
Answer:
f(3) = 9
Step-by-step explanation:
f(x) = x^2
f(3) means "When x is 3" So, we substitute x with 3.
f(3) = 3^2
f(3) = 3 x 3 = 9
f(3) = 9
