Answer:
b= 1.1
Step-by-step explanation:
193b= 212.3
divide both sides by 193 to isolate b
212.3÷193=1.1
Answer:
1+1= 2
Step-by-step explanation:
or sometimes people say 11, but irl its 2
Answer:
- Part A: The price of fuel A is decreasing by 12% per month.
- Part B: Fuel A recorded a greater percentage change in price over the previous month.
Explanation:
<u>Part A:</u>
The function 
calculates the price of fuel A each month by multiplying the price of the month before by 0.88.
Month price, f(x)
1 2.27 (0.88) = 1.9976 ≈ 2.00
2 2.27(0.88)² = 1.59808 ≈ 1.60
3 2.27(0.88)³ = 1.46063 ≈ 1.46
Then, the price of fuel A is decreasing.
The percentage per month is (1 - 0.88) × 100 = 12%, i.e. the price decreasing by 12% per month.
<u>Part B.</u>
<u>Table:</u>
m price, g(m)
1 3.44
2 3.30
3 3.17
4 3.04
To find if the function decreases with a constant ration divide each pair con consecutive prices:
- ratio = 3.30 / 3. 44 = 0.959 ≈ 0.96
- ratio = 3.17 / 3.30 = 0.960 ≈ 0.96
- ratio = 3.04 / 3.17 = 0.959 ≈ 0.96
Thus, the price of fuel B is decreasing by (1 - 0.96) × 100 =4%.
Hence, the fuel A recorded a greater percentage change in price over the previous month.
Answer:
p is approximately equal to 1.125, -1.925
Step-by-step explanation:
We can solve this first by expanding, giving us a quadratic equation in the usual ax² + bx + c format, then solving that for p:
(4p - 4)(13p + 27) = 0
52p² + 108p - 52p - 108 = 0
52p² + 52p - 108 = 0
13p² + 13p - 27 = 0
13p² + 13p = 27
p² + p = 27 / 13
p² + p + 1/4 = 27/13 + 1/4
(p + 1/2)² = 108 / 52 + 13 / 52
(p + 1/2)² = 121 / 52
p + 1/2 = ± √(121 / 52)
p = -1/2 ± 11√(1/13) / 2
At that point we can simply plug those numbers into a calculator and solve it:
p ≈ 1.125, -1.925
Answer: 23F - (-8F)
Step-by-step explanation:
We have been given that at midnight, the temperature was . At noon, the temperature was .
Since our temperature has increased to 23 degree Fahrenheit from -8 degree Fahrenheit, we can represent this increase in temperature as:
Therefore, the expression represents the increase in temperature.
