This is a problem you need to solve using logs. When you use logs you can "pull" the exponents down in front of the log to get a new equation that looks like this: 2x^3 + x^2 log 81 = 6x - 3 log 27. Now divide both sides by log 81 and 6x - 3 simultaneously to get (2x^3 + x^2)/(6x - 3) = (log 27)/(log 81). If you do the log math on the right side you get .75. Now multiply both sides by 6x-3 to get 2x^3+x^2 = .75(6x-3). If you distribute that out on the left side you'll get 2x^3+x^2=4.5x-2.25. Now move everything over to the left side and set the whole thing equal to 0: 2x^3+x^2-4.5x+2.25=0. When you solve for x, you are in essence factoring, so do this by grouping: x^2(2x+1)-2.25(2x+1). Now finally factor out the 2x+1 to get (2x+1)(x^2-2.25). You're not done yet though cuz you need to solve each of those for x: 2x+1=0, and x= -1/2; x^2=2.25, and x=+/- 1.5. So all the values for x here are -1/2, 1.5, and -1.5
Answer:
The correct option is b.
Step-by-step explanation:
The formula for standard deviation is
where, 
is mean of the data and n is number of observation.
The variance of a stock's returns can be calculated by the above formula.
Variance of stock's returns is the average value of squared deviations from the mean.
Therefore the correct option is b.
Answer:
x = - 8, x = 6
Step-by-step explanation:
Given the 2 equations
x² + y² = 100 → (1)
y = x + 2 → (2)
Substitute y = x + 2 into (1)
x² + (x + 2)² = 100
x² + x² + 4x + 4 = 100
2x² + 4x + 4 = 100 ( subtract 100 from both sides )
2x² + 4x - 96 = 0 ( divide through by 2 )
x² + 2x - 48 = 0 ← in standard form
(x + 8)(x - 6) = 0 ← in factored form
Equate each factor to zero and solve for x
x + 8 = 0 ⇒ x = - 8
x - 6 = 0 ⇒ x = 6
<span>Tabia spent 15/24 of this week's allowance on gummy bears. She can calculate that by multiplying 3/4 times 5/6. To do that, she multiplies the top numbers of each fraction together (3 times 5 equals 15), and to get the top number of the fraction. Then she multiplies the bottom numbers together (4 times 6 equals 24) to get the bottom number.</span>
