Answer:
B. b = 3a + 2
Step-by-step explanation:
We can write the equation in slope-intercept form as b = ma + c, where,
m = slope/rate of change
c = y-intercept/initial value
✔️Find m using any two given pair of values, say (2, 8) and (4, 14):
Rate of change (m) = change in b/change in a
m = (14 - 8)/(4 - 2)
m = 6/2
m = 3
✔️Find c by substituting (a, b) = (2, 8) and m = 3 into b = ma + c. Thus:
8 = 3(2) + c
8 = 6 + c
8 - 6 = c
2 = c
c = 2
✔️Write the equation by substituting m = 3 and c = 2 into b = ma + c. Thus:
b = 3a + 2
400$
Explanation:
In 1992, the price is 28000$ and 9600$ in 1996 so in 4 years the value the car lost is 18400$
Dividing 18400 by 4 years we get 4600 depreciation value per year.
Between 1996 and 1998 there is 2 years
So in 1998 the value of the car will be:
9600 (value in 1996) - 2x4600 = 400$
Answer:
g(x) = log2 (x) – 1
Step-by-step explanation:
I just did this on Edge
Answer:
a) r₁₂ = 104.36
In general, rₙ = arⁿ⁻¹
b)
- rabbit food consumed during the 10th year is approximately 832 pounds
- rabbit food consumed in total for the 1st through 10th years is approximately 5265 pounds
Step-by-step explanation:
Given that:
r1 = 30 and a farm grows by 12%
a = 30 and the common ratio r = 1.12
now
n r
1 30.00
2 33.60
3 37.63
4 42.15
5 47.21
6 52.87
7 59.21
8 66.32
9 74.28
10 83.19
11 93.18
12 104.36
Therefore r₁₂ = 104.36
In general, rₙ = arⁿ⁻¹
b)
if each rabbit consume 10 lbs of rabbit food each year
n r food consumed(lbs)
1 30.00 300
2 33.60 336
3 37.63 376
4 42.15 422
5 47.21 472
6 52.87 529
7 59.21 592
8 66.32 663
9 74.28 743
10 83.19 832
total 5265
Therefore, the rabbit food consumed during the 10th year is approximately 832 pounds
And the rabbit food consumed in total for the 1st through 10th years is approximately 5265 pounds
