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2 years ago

Help Please really do need help

Mathematics

1 answer:

Ksju [112]2 years ago

5 0

Answer:

your answer would be a whole number, so your answer would be 1,000 i think.

this should be the answer because if you multiply

9*1000 you get 9000

if you multiply

25 * 1000 you get 25000

so on and so on

Step-by-step explanation:

hope this helps :)

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The annual net income of a company for the period 2007–2011 could be approximated by P(t) = 1.6t2 − 11t + 44 billion dollars (2

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Answer:

$P'(t) = 3.2 t -11$

And we can set equal this derivate to 0 in order to find the critical point and we got:

$3.2 t-11= 0$

$t = \frac{11}{3.2}= 3.4375$

And we can calculate the second derivate and we got:

$P''(t) = 3.2 >0$

So then w can conclude that the value of t = 3.4375 represent the minimum value for the function and we can replace in the original function and we got:

$P(3.4375) = 1.6(3.4375)^2 - (11*3.4375) +44 = 25.094$

So then the minimum annual income occurs at t = 3.43 (between 2008 and 2009) and the value is 25.094

Step-by-step explanation:

For this case we have the following function:

$P(t) = 1.6 t^2 -11t +44$

Where P represent the annual net income for the period 2007-2011 and $2 \leq t \leq 7$

And t represent the time in years since the start of 2005

In order to find the lowet income we need to use the derivate, given by:

$P'(t) = 3.2 t -11$

And we can set equal this derivate to 0 in order to find the critical point and we got:

$3.2 t-11= 0$

$t = \frac{11}{3.2}= 3.4375$

And we can calculate the second derivate and we got:

$P''(t) = 3.2 >0$

So then w can conclude that the value of t = 3.4375 represent the minimum value for the function and we can replace in the original function and we got:

$P(3.4375) = 1.6(3.4375)^2 - (11*3.4375) +44 = 25.094$

So then the minimum annual income occurs at t = 3.43 (between 2008 and 2009) and the value is 25.094

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