Answer:
Option C. f(n) = 16(3/2)⁽ⁿ¯¹⁾
Step-by-step explanation:
To know which option is correct, do the following:
For Option A
f(n) = 3/2(n – 1) + 16
n = 1
f(n) = 3/2(1 – 1) + 16
f(n) = 3/2(0) + 16
f(n) = 16
n = 2
f(n) = 3/2(n – 1) + 16
f(n) = 3/2(2 – 1) + 16
f(n) = 3/2(1) + 16
f(n) = 3/2 + 16
f(n) = 1.5 + 16
f(n) = 17.5
For Option B
f(n) = 3/2(16)⁽ⁿ¯¹⁾
n = 1
f(n) = 3/2(16)⁽¹¯¹⁾
f(n) = 3/2(16)⁰
f(n) = 3/2 × 1
f(n) = 1
For Option C
f(n) = 16(3/2)⁽ⁿ¯¹⁾
n = 1
f(n) = 16(3/2)⁽¹¯¹⁾
f(n) = 16(3/2)⁰
f(n) = 16 × 1
f(n) = 16
n = 2
f(n) = 16(3/2)⁽ⁿ¯¹⁾
f(n) = 16(3/2)⁽²¯¹⁾
f(n) = 16(3/2)¹
f(n) = 16(3/2)
f(n) = 8 × 3
f(n) = 24
n = 3
f(n) = 16(3/2)⁽ⁿ¯¹⁾
f(n) = 16(3/2)⁽³¯¹⁾
f(n) = 16(3/2)²
f(n) = 16(9/4)
f(n) = 4 × 9
f(n) = 36
For Option D
f(n) = 8n + 8
n = 1
f(n) = 8(1) + 8
f(n) = 8 + 8
f(n) = 16
n = 2
f(n) = 8n + 8
f(n) = 8(2) + 8
f(n) = 16 + 8
f(n) = 24
n = 3
f(n) = 8n + 8
f(n) = 8(3) + 8
f(n) = 24 + 8
f(n) = 32
From the above illustration, only option C describes the sequence.
So, the problem is asking for one of the equations to be rewritten. Lets rewrite the second one. The way to create a system of equations that will cancel is to create a situation where the a or b of both equations add to 0. In this case, we can multiply the second equation by 3 to turn the 3y into 9y.
Our new system is 
and . We can now add these together.
Adding each part straight down, we get the equation:
We now have an equation with a single variable, which also tells us that x is equal to 27.
Extra: To find the value of y after this, we merely need to plug in the value of x to one of the equations and solve for y.
1. 4x=64
x=26
2. 9x=20
x=9/20
3. 8+5x<70
5x<62
x<12.4
4. 2+x=50
x=52
5. n+5-5<15
n<15
6. x-10=30
x=40
7. x+5=13
x=8
8. a+8=20
a=12
9. x·3=27
x=9
10. x²=2+6
x²=8
x=√8
x=-√8
Answer:
You should probably be more specific as different schools have different curriculum's!
Step-by-step explanation:
Answer:
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Step-by-step explanation:
