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:
2πr
Step-by-step explanation:
This is the concept of algebra, the solution of the given expression will be given as follows;
-5x+2x²=-6x
adding 5x on both sides we get
5x-5x+2x²=-6x+5x
2x²=-x
divide both sides by 2
(2x²)/2=-x/2
x²=-x/2
divide both sides by x
(x²)/x=-x/(2x)
x=-1/2
the answer is x=-1/2
Answer:
15
Step-by-step explanation:
The difference between consecutive integers is 1.
The difference between consecutive odd integers is 2.
Let the smallest odd integer be x.
Then the next greater one is x + 2. The greatest one is x + 4.
"3 times the first" is 3x
"twice the third" is 2(x + 4)
"3 times the first of three consecutive odd integers is 3 more than twice the third"
3x = 2(x + 4) + 3
3x = 2x + 8 + 3
x = 11
The smallest integer is 11.
x + 4 = 11 + 4 = 15
The greatest one is 15.
Answer: 74
Step-by-step explanation:
nth term= 6n+2 (work out the nth term)
6x12+2=74 (substitute 12 in (to find the 12th term))
12th term= 74
