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3 years ago

HELP DUE IN 30 MINS!

Mathematics

1 answer:

ira [324]3 years ago

3 0

Answer:

Arc Length = 65.9 ft

Step-by-step explanation:

Arc Length = 2π(r) * Central Angle/360

=> Arc Length = 2π(14) * 270/360

=> Arc Length = 87.96 * 0.75

=> Arc Length = 65.9 ft

Hope this helps!

Thx for unfriending me btw :)

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Fofino [41]

Answer:

Cool. Thanks!

Step-by-step explanation:

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3 years ago

Read 2 more answers

All boxes with a square base, an open top, and a volume of 60 ft cubed have a surface area given by S(x)equalsx squared plus

Karo-lina-s [1.5K]

Answer:

The absolute minimum of the surface area function on the interval $(0,\infty)$ is $S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

The dimensions of the box with minimum surface area are: the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

Step-by-step explanation:

We are given the surface area of a box $S(x)=x^2+\frac{240}{x}$ where x is the length of the sides of the base.

Our goal is to find the absolute minimum of the the surface area function on the interval $(0,\infty)$ and the dimensions of the box with minimum surface area.

1. To find the absolute minimum you must find the derivative of the surface area ( $S'(x)$ ) and find the critical points of the derivative ( $S'(x)=0$ ).

$\frac{d}{dx} S(x)=\frac{d}{dx}(x^2+\frac{240}{x})\\\\\frac{d}{dx} S(x)=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(\frac{240}{x}\right)\\\\S'(x)=2x-\frac{240}{x^2}$

Next,

$2x-\frac{240}{x^2}=0\\2xx^2-\frac{240}{x^2}x^2=0\cdot \:x^2\\2x^3-240=0\\x^3=120$

There is a undefined solution $x=0$ and a real solution $x=2\sqrt[3]{15}$ . These point divide the number line into two intervals $(0,2\sqrt[3]{15})$ and $(2\sqrt[3]{15}, \infty)$

Evaluate S'(x) at each interval to see if it's positive or negative on that interval.

$\begin{array}{cccc}Interval&x-value&S'(x)&Verdict\\(0,2\sqrt[3]{15}) &2&-56&decreasing\\(2\sqrt[3]{15}, \infty)&6&\frac{16}{3}&increasing \end{array}$

An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We can see from the table that f(x) decreases before $x=2\sqrt[3]{15}$ , increases after it, and is defined at $x=2\sqrt[3]{15}$ . So f(x) has a relative minimum point at $x=2\sqrt[3]{15}$ .

To confirm that this is the point of an absolute minimum we need to find the second derivative of the surface area and show that is positive for $x=2\sqrt[3]{15}$ .

$\frac{d}{dx} S'(x)=\frac{d}{dx}(2x-\frac{240}{x^2})\\\\S''(x) =\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(\frac{240}{x^2}\right)\\\\S''(x) =2+\frac{480}{x^3}$

and for $x=2\sqrt[3]{15}$ we get:

$2+\frac{480}{\left(2\sqrt[3]{15}\right)^3}\\\\\frac{480}{\left(2\sqrt[3]{15}\right)^3}=2^2\\\\2+4=6>0$

Therefore S(x) has a minimum at $x=2\sqrt[3]{15}$ which is:

$S(2\sqrt[3]{15})=(2\sqrt[3]{15})^2+\frac{240}{2\sqrt[3]{15}} \\\\2^2\cdot \:15^{\frac{2}{3}}+2^3\cdot \:15^{\frac{2}{3}}\\\\4\cdot \:15^{\frac{2}{3}}+8\cdot \:15^{\frac{2}{3}}\\\\S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

2. To find the third dimension of the box with minimum surface area:

We know that the volume is 60 $ft^3$ and the volume of a box with a square base is $V=x^2h$ , we solve for h

$h=\frac{V}{x^2}$

Substituting V = 60 $ft^3$ and $x=2\sqrt[3]{15}$

$h=\frac{60}{(2\sqrt[3]{15})^2}\\\\h=\frac{60}{2^2\cdot \:15^{\frac{2}{3}}}\\\\h=\sqrt[3]{15} \:ft$

The dimension are the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

6 0

3 years ago

p Mes mm place .5 35 yards long She needs to cut frees 3 in it s we greatest number of pieces that she can ea A

bonufazy [111]

She will have aprox. 40 pieces of ribbon

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3 years ago

A cheerleading squad consists of ten cheerleaders of ten different heights. How many ways are there for the cheerleaders to line

Sindrei [870]

Answer:

11340

Step-by-step explanation:

We have 10 cheerleaders of ten different heights. There are two rows each having 5 positions so that five cheerleaders occupy the first row that have height less than the five cheerleaders who occupied the second row.

Therefore, Out of ten cheerleaders, two cheerleaders will be selected out of which the tallest one will occupy the position in the upper row and the second one will occupy in the lower row, therefore these two positions will be occupied in $10C_{2}$ ways. Now, we are left with 8 cheerleaders and following the same path, the next two positions will also be occupied in $8C_{2}$ ways, similarly, following the same path, the number of ways the cheerleaders will occupy the position will be: $10C_{2}$ × $8C_{2}$ × $6C_{2}$ × $4C_{2}$ × $2C_{2}$

=11340

8 0

3 years ago

PLEASE HELP! THANKS!

Vinvika [58]

So that is area of rectangle-1/2areacircle

so
areacircle=pir^2
r=5
1/2pir^2=1/2pi5^2=25pi/2=12.5pi

arearectangle=lw=12*5=60

unshaded=60-12.5pi=20.75 square feet
round

20.8 square feet

6 0

3 years ago