I'll do problem 1 to get you started.
The vertical sides are 3 and x for the left and right figures.
The horizontal sides are 15 and 60 for the left and right figures.
The corresponding sides form fractions which are equal (due to the nature of the similar polygons)
3/x = 15/60
3*60 = x*15 ... cross multiply
180 = 15x
15x = 180
x = 180/15 .... divide both sides by 15
x = 12
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Another way you can see this is note how the jump from 15 to 60 is "times 4", so the jump from 3 to x must also be "times 4" to keep the same proportion
3 ---> x = 3*4 = 12
Or you could set up the proportion
60/15 = x/3
4 = x/3
x/3 = 4
x = 3*4
x = 12
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<h3>Final Answer: 12</h3>
X+y= 26 and 4.75x + 2.25y= 83.50
y=26-x( substitute this into second equation for y and solve for x)
4.75x + 2.25(26-x)= 83.50
4.75x + 58.5-2.25x= 83.50
x= 10
Now solve for y by substituting your answer for x
10+ y= 26
y=16
Therefore, 16 tickets were purchased for kids and 10 for adults.
Answer:
Tn = 2Tn-1 - Tn-2
Step-by-step explanation:
Before we can generate the recursive sequence, we need to find the nth term of the given sequence.
nth term of an AP is given as:
Tn = a+(n-1)d
If a17 = -40
T17 = a+(17-1)d = -40
a+16d = -40 ...(1)
If a28 = -73
T28 = a+(28-1)d = -73
a+27d = -73 ...(2)
Solving both equations simultaneously using elimination method.
Subtracting 1 from 2 we have:
27d - 16d = -73-(-40)
11d = -73+40
11d = -33
d = -3
Substituting d = -3 into 1
a+16(-3) = -40
a - 48 = -40
a = -40+48
a = 8
Given a = 8, d = -3, the nth term of the sequence will be
Tn = 8+(n-1) (-3)
Tn = 8+(-3n+3)
Tn = 8-3n+3
Tn = 11-3n
Given Tn = 11-3n and d = -3
Tn-1 = Tn - d... (3)
Tn-1 = 11-3n +3
Tn-1 = 14-3n
Tn-2 = Tn-2d...(4)
Tn-2 = 11-3n-2(-3)
Tn-2 = 11-3n+6
Tn-2 = 17-3n
From 3, d = Tn - Tn-1
From 4, d = (Tn - Tn-2)/2
Equating both common difference
(Tn - Tn-2)/2 = Tn - Tn-1
Tn - Tn-2 = 2(Tn - Tn-1)
Tn - Tn-2 = 2Tn-2Tn-1
2Tn-Tn = 2Tn-1 - Tn-2
Tn = 2Tn-1 - Tn-2
The recursive formula will be
Tn = 2Tn-1 - Tn-2
Answer:
It is linear
Step-by-step explanation:
The Y goes up by the same amount as the X (by 4 Y's per 1 X)
Okay here we go!
1. I'm a visual person so i would start by drawing circles for the wheels.
o o o o o o o o o o o o o o o o o o o
2. I tried separating them by threes
ooo ooo ooo ooo ooo ooo o
2 By two's
oo oo oo oo oo oo oo oo oo o
3. then i started to count
ooo ooo ooo ooo ooo = 15 and i still had 4 left over
oo oo = 4 and that would equal to 19 wheels!
the answer is: there are 5 trikes and 2 bicycles!!!!!
Hope that helped!
:)
