Steps:
1. Change 1/11 to 2/22 by multiplying by 2
2. 9/22 + 2/22
3. 11/22
Then, to simplify it you divide by 11
4. Your answer is
1/2
I dont know if this is what you are looking for but i think this is the answer of the question you are looking for which is the general equation of a horizontal hyperbola:
<span>(x-h)²/a² - (y-k)²/b² = 1 </span>
<span>with </span>
<span>center (h,k)
</span>What you need to do is replace the data and do the calculations
Answer:
Step-by-step explanation:
let y represent number of years...
9 ft tall = (12 * 9) = 108 inches
10 ft tall = (12 * 10) = 120 inches
Type A : 9 ft (or 108 inches) and grows 6 inches per year..
108 + 6y
Type B : 10 ft (or 120 inches) and grows 3 inches per yr
120 + 3y
how many years will it take them to be the same height...so when will they be equal....so set them equal and solve for y
108 + 6y = 120 + 3y
6y - 3y = 120 - 108
3y = 112
y = 112/3
y = 37 1/3 years....or 37 years and 4 months <====
check...
108 + 6(37 1/3) = 120 + 3(37 1/3)
108 + 6(112/3) = 120 + 3(112/3)
108 + 672/3 = 120 + 336/3
672/3 - 336/3 = 120 - 108
336/3 = 112
112 = 112 (correct)
Answer:
A
Step-by-step explanation:
If the number of people in the theater must be under 240, then there must be less than 10 seats in each row on the floor to meet fire safety regulations.
if X=10 then, 15 times 10 plus 90 = 240.
Since the number of people must be under 240, there must be less than 10 seats in each row on the floor.
Answer:
4.62 years
Step-by-step explanation:
Half the car's value =>$24,000/2 = $12,000
The formula for exponential decrease =
y = a(1 - r)^t
y = Initial value = $24,000
a = Value after time t = $12,000
r = Decrease rate = 15% = 0.15
t = time in years =??
Hence,
24,000 = 12000 ( 1 - 0.15)^t
Divide both sides by 12000
24000/12000 = 12000 ( 0.85)^t/ 12000
2 = 0.85^t
We take the logarithm of both sides
log 2 = log (0.85)^t
log 2 = t log 0.85
Divide both sides by log 0.85
t = log 2/log 0.85
t = 4.62098120373 years
Approximately = 4.62 years
It would take 4.62 years to be worth half the initial price
