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2 years ago

(12y²+17y-4)+(9y²-13y+3)

Mathematics

1 answer:

PSYCHO15rus [73]2 years ago

5 0

(12y^2 + 17y - 4) + (9y^2 - 13y + 3)

= 12y^2 + 17y - 4 + 9y^2 - 13y + 3

= 21y^2 + 4y - 1

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3x+3

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3 years ago

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 $h=\sqrt[3]{15} \:ft$

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3 years ago

URGENT!!LAST QUESTION!!!! PLS JUST TAKE A LOOK! EASY BUT I AM DUMB!!!!!! WILL GIVE BRANLIEST!!

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Answer:

x = 28

Step-by-step explanation:

If the quadrilaterals are similar, there is a proportionality among their sides:

The top side in the large figure (70) is to the top side in the small figure (10) in the same ratio as the left side (x) in the large figure is to the left side in the small figure (4). This in math terms is written as:

$\frac{70}{10} =\frac{x}{4}$

We can then solve for the unknown "x" by multiplying both sides by 4:

$\frac{70}{10} =\frac{x}{4}\\\frac{70\,*\,4}{10} =x\\x=28$

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3 years ago

What is the sum of 1 2/5 and 5 3/4

Art [367]

Answer:

$7\frac{3}{20}$

Step-by-step explanation:

Hey there!

Well to add this we need to pu it in improper form.

7/5 + 23/4

Now we need to find the LCM.

5 - 5, 10, 15, 20, 25, 30

4 - 4, 8, 12, 16, 20, 24, 28

So the LCD is 20.

Now we need to change the 5 and 4 to 20.

5*4 = 20

7*4 = 28

28/20

4*5=20

23*5=115

115/20

Now we can add 28 and 115,

= 143/20

Simplified

7 3/20

Hope this helps :)

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3 years ago

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(12y²+17y-4)+(9y²-13y+3)​

(12y²+17y-4)+(9y²-13y+3)