Answer:
c is the correct option
Step-by-step explanation:
from,
f'(x) = h >0 <u>f</u><u>(</u><u>x</u><u> </u><u>+</u><u> </u><u>h</u><u>)</u><u> </u><u>-</u><u> </u><u>f</u><u>(</u><u>x</u><u>)</u><u> </u>
h
f(x) = - √2x
f(x + h) = - √(2x + h)
f'(x) = h>0 <u>-</u><u>√(2x + h) - √2x</u>
h
rationalize the denominator
= h>0 <u>-</u><u>√</u><u>(</u><u>2</u><u>x</u><u> </u><u>+</u><u> </u><u>h</u><u>)</u><u> </u><u>+</u><u> </u><u>√</u><u>2</u><u>x</u><u> </u><u> </u><u>(</u><u>-</u><u>√</u><u>(</u><u>2</u><u>x</u><u> </u><u>+</u><u> </u><u>h</u><u>)</u><u> </u><u>-</u><u> </u><u>√</u><u>2</u><u>x</u><u>)</u>
h (-√(2x + h) - √2x)
= h>0 <u>4</u><u>x</u><u> </u><u>+</u><u> </u><u>2</u><u>h</u><u> </u><u>-</u><u> </u><u>4</u><u>x</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>
h(-√(2x + h) -√2x)
= h>0 <u>2</u><u>h</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>
h(-√(2x+h) - √2x)
= h>0 <u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>2</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>
-√(2x+h) - √2x
Answer:
end of third year = 1,200*(1.02)^3 = 1,273.45 plus
end of second year = 1,200*(1.02)^2 = 1,248.48 plus
end of first year = 1,200*(1.02) = 1,224.00
Total = 3,745.9
Step-by-step explanation:
The ratio would be 14:6
If you reduce it it would be 7:3
Since all of the bases (x) are equal you just add the exponents.
(The x has an exponent of 1)
3 + -4 + 1
-1 + 1
0
Since the answer is 0 it is x^0
And any number or variable set to the power of 0 equals 1, therefore...
ANSWER: x^3 • x^-4 • x = 1
Hope this helps!
The answer would be C) P,D and A