it is the same denominator therefore only the numerators are added and the sign is maintained. The answer is -12/9 but it can be reduced to -4/3.
Answer:
21
Step-by-step explanation:
Answer:
Part a)
Part b)
Step-by-step explanation:
we know that
The formula to calculate continuously compounded interest is equal to
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
e is the mathematical constant number
Part a) What is the amount after 2 years?
we have
substitute in the formula above
Part b) How long will it take for the amount to be $8000?
we have
substitute in the formula above and solve for t
Applying ln both sides
Remember that
ln(e)=1
No one can help you because you didn't provide us with anything
Use the chain rule:
<em>y</em> = tan(<em>x</em> ² - 5<em>x</em> + 6)
<em>y'</em> = sec²(<em>x</em> ² - 5<em>x</em> + 6) × (<em>x</em> ² - 5<em>x</em> + 6)'
<em>y'</em> = (2<em>x</em> - 5) sec²(<em>x</em> ² - 5<em>x</em> + 6)
Perhaps more explicitly: let <em>u(x)</em> = <em>x</em> ² - 5<em>x</em> + 6, so that
<em>y(x)</em> = tan(<em>x</em> ² - 5<em>x</em> + 6) → <em>y(u(x))</em> = tan(<em>u(x)</em> )
By the chain rule,
<em>y'(x)</em> = <em>y'(u(x))</em> × <em>u'(x)</em>
and we have
<em>y(u)</em> = tan(<em>u</em>) → <em>y'(u)</em> = sec²(<em>u</em>)
<em>u(x)</em> = <em>x</em> ² - 5<em>x</em> + 6 → <em>u'(x)</em> = 2<em>x</em> - 5
Then
<em>y'(x)</em> = (2<em>x</em> - 5) sec²(<em>u</em>)
or
<em>y'(x)</em> = (2<em>x</em> - 5) sec²(<em>x</em> ² - 5<em>x</em> + 6)
as we found earlier.