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2 years ago

An initial investment of $3000 compounded annually at a rate of 12% after 20 years.

Mathematics

1 answer:

Doss [256]2 years ago

5 0

Answer:

$=\frac{2^{\frac{2}{3}}\sqrt[3]{x}}{2x}$

Step-by-step explanation:

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The Davidson family wants to expand its rectangular patio, which currently measures 15 ft by 12 ft. They want to extend the leng

tiny-mole [99]

You just need to do 15x12=180+160=340divided by 2

8 0

2 years ago

Read 2 more answers

James is the manager at an entertainment arena that draws an average 7,000 patrons per event. Each ticket taker can process 350

Elina [12.6K]

Answer:

No. James didn't have enough ticket takers to process the average number of patrons hat usually attend the events.

He need to hire 2 more ticket taker (i.e 20 ticket takers) in order to process the average number of patrons hat usually attend the events.

Step-by-step explanation:

Given:

Average amount of patron per event= 7000

Each ticket taker can process = 350

Number of ticket takers hired = 18

We need to find the whether he hire enough to process the average number of patrons that usually attend the events.

Solution:

We will find the Average amount of patron ticket takers can process.

Now we can say that

Average amount of patron ticket takers can process is equal to Each ticket taker can process multiplied by Number of ticket takers hired.

framing in equation form we get;

Average amount of patron ticket takers can process = $350 \times 18 = 6300\ patrons$

Hence With the required hires James cannot fully process the patrons usually attending the event.

To find how many ticket takers required we will divide average number of patrons with Each ticket taker can process.

framing in equation form we get;

ticket takers required = $\frac{7000}{350}=20$

hence In order t process the the average number of patrons that usually attend the events James will require 20 ticket takers.

6 0

2 years ago

Use lagrange multipliers to find the points on the given surface that are closest to the origin. y2 = 64 + xz

Inessa [10]

The distance between an arbitrary point on the surface and the origin is

$d(x,y,z)=\sqrt{x^2+y^2+z^2}$

Recall that for differentiable functions $g(x)$ and $h(x)$ , the composition $g(h(x))$ attains extrema at the same points that $h(x)$ does, so we can consider an augmented distance function

$D(x,y,z)=x^2+y^2+z^2$

The Lagrangian would then be

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(y^2-64-xz)$

We have partial derivatives

$\begin{cases}L_x=2x-\lambda z\\L_y=2y+2y\lambda\\L_z=2z-\lambda x\\L_\lambda=y^2-64-xz\end{cases}$

Set each partial derivative to 0 and solve the system to find the critical points.

From the second equation it follows that either $y=0$ or $\lambda=-1$ . In the first case we arrive at a contradiction (I'll leave establishing that to you). If $\lambda=-1$ , then we have

$\begin{cases}2x+z=0\\2z+x=0\end{cases}\implies x=0,z=0$

This means $y^2=64\implies y=\pm8$

so that the points on the surface closest to the origin are $(0,\pm8,0)$ .