For domain 2x sqrt(2+x)>0
x>0,2+x>0,x>-2 combining
we get x>2
f'(x)=[1/{2x sqrt(2+x)}][{2x/(2 sqrt(2+x))}+2 sqrt(2+x)]
Answer: 
Step-by-step explanation:
The first step is to make the division of the fractions 
and
. To do this, you can flip the fraction over and multiply the numerators and the denominators of the fractions. Then:
Reduce the fraction 
:
Now you can make the subtraction: in this case the Least Common Denominator (LCD) will be the multiplication of the denominators. Divide each denominator by the LCD and multiply this quotient by the corresponding numerator and then subtract the products. Therefore you get:
Answer:x=11
Step-by-step explanation:
2x-13=9
2x=9+13
2x=22
x=22/2
x=11
hope this helps :)
Answer:
a) x1 = 6 and x2 = -2
b) -2
Step-by-step explanation:
a)
To find the roots of the quadratic equation, we can use the Bhaskara's formula:
Delta = b^2 - 4ac
Delta = (-4)^2 - 4*1*(-12) = 64
sqrt(Delta) = 8
x1 = (-b + sqrt(Delta)) / 2a
x1 = (4 + 8) / 2
x1 = 6
x2 = (-b - sqrt(Delta)) / 2a
x2 = (4 - 8) / 2
x2 = -2
b)
The roots are 6 and -2, so the smaller root is -2
You just have to divide the decimals
