Three consecutive numbers are x, x+1 and x+2.
Four times the first integer is 4x
The sum of the second and third is (x+1)+(x+2)=2x+3.
So, we have
Subtract 2x from both sides:
Divide both sides by 2:
So, you can't have three consecutive integers such that four times the first is 18 more than the sum of the other two: the three numbers would be 10.5, 11.5, 12.5.
In fact, you have
and
Use the rules of logarithm:
1. log(x)+log(y)=log(xy)
log(x)-log(y)=log(x/y)
2. k*log(x) = log(x^k)
log(x)/k = log(x^(1/k))
<span>log(2z)+2log(2x)+4log(9y)+12log(9x)−2log(2y)
=</span>log(2z)+log(4x^2)+log(9^4y^4)+log(9^12x^12)−log(4y^2)
=log(2z)+log(4x^2)+log(6561y^4)+log(282429536481x^12)−log(4y^2)
=log(59296646043258912 * x^14 * y^6 * z)
You have to use the equation (72+y)=5(4+y) which Y =+13
Answer:
12x + 20
Step-by-step explanation:
2(4x + 6 +2x + 4)
Answer:
5.40 is not the same as 7.00
Step-by-step explanation:
