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3 years ago

Consider this Revenue Advertising Expense situation.

Mathematics

2 answers:

Dmitrij [34]3 years ago

7 0

Answer:

C. Package C: Php 32,000.00

Step-by-step explanation:

<u>Given function</u>

P = −50x³ + 2400x² − 2000, 0 ≤ x ≤ 32

x is the amount spent on advertising (in thousands of pesos).

Substitute x with 8, 16 and 32 and solve for P.

<u>The greatest value of P is at: </u>

x = 32 ( Php 32000.00)

and the value is

P = 817200.00

Another method is graphing.

See the attached picture for graphical solution.

Serhud [2]3 years ago

4 0

Answer:

Package C: Php 32,000.00

Step-by-step explanation:

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Solve. 12 students in the classroom and she wants to create a student to teacher ratio of three to one (three students per teach

Natasha2012 [34]

She will need a one to one ratio so to get to your ratio you need to take your 3 to 1 ratio and divide it by the 1 on 1 ratio

7 0

3 years ago

The midpoint of CD is M=(1, 2). One endpoint is C = (6, -2).

Zina [86]

Answer: Endpoint = (-4,6)

Step-by-step explanation:

To find any endpoint use the formula $\frac{x_1+x_2}{2} = Xmp$ $\frac{y_1+y_2}{2} =Ymp$

$\frac{6+x_2}{2} = 1$ $\frac{-2+y_2}{2} =2$

multiply each by 2 to simplify

$\frac{6+x_2}{2*2} = 1*2 = {6+x_2}= 2$ $\frac{-2+y_2}{2*2} =2*2 = -2+y_2=4$ take the equations and simplify

6+ $x_{2}$ = 2 -2+ $y_{2}$ = 4 =

-6 -6 +2 +2

$x_{2}$ = -4 $y_{2}$ = 6

To check use the Midpoint formula

$\frac{-4+6}{2} , \frac{6-2}{2} = \frac{2}{2} , \frac{4}{2} = (1,2)$

8 0

3 years ago

Find the sum of the geometric series 512+256+ . . .+4

mario62 [17]

$\bf 512~~,~~\stackrel{512\cdot \frac{1}{2}}{256}~~,~~...4$

so, as you can see above, the common ratio r = 1/2, now, what term is +4 anyway?

$\bf n^{th}\textit{ term of a geometric sequence}\\\\a_n=a_1\cdot r^{n-1}\qquad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\r=\frac{1}{2}\\a_1=512\\a_n=+4\end{cases}$

$\bf 4=512\left( \cfrac{1}{2} \right)^{n-1}\implies \cfrac{4}{512}=\left( \cfrac{1}{2} \right)^{n-1}\\\\\\\cfrac{1}{128}=\left( \cfrac{1}{2} \right)^{n-1}\implies \cfrac{1}{2^7}=\left( \cfrac{1}{2} \right)^{n-1}\implies 2^{-7}=\left( 2^{-1}\right)^{n-1}\\\\\\(2^{-1})^7=(2^{-1})^{n-1}\implies 7=n-1\implies \boxed{8=n}$

so is the 8th term, then, let's find the Sum of the first 8 terms.

$\bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\r=\frac{1}{2}\\a_1=512\\n=8\end{cases}$

$\bf S_8=512\left[ \cfrac{1-\left( \frac{1}{2} \right)^8}{1-\frac{1}{2}} \right]\implies S_8=512\left(\cfrac{1-\frac{1}{256}}{\frac{1}{2}} \right)\implies S_8=512\left(\cfrac{\frac{255}{256}}{\frac{1}{2}} \right)\\\\\\S_8=512\cdot \cfrac{255}{128}\implies S_8=1020$