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3 years ago

Help I know is easy but I don't know

Mathematics

2 answers:

monitta3 years ago

8 0

Answer:

$? =9.16515139$

Step-by-step explanation:

→You can use the Pythagorean Theorem:

$a^2+b^2=c^2$

→The "a," and "b," are the legs, and the "c," is the hypotenuse. In this case, we area already given the hypotenuse (10), and are solving for the leg. So plug in the numbers, like so:

$4^2+b^2=10^2$

$16 + b^2=100$

→Subtract 16 from both sides:

$b^2 = 84$

→To get "b," by itself, square root both sides, like so:

$\sqrt{b^2} = \sqrt{84}$

$b = 9.16515139$

n200080 [17]3 years ago

4 0

Answer:

? = 2 sqrt(21)

Step-by-step explanation:

We can use the Pythagorean theorem to solve since this is a right triangle

a^2 + b^2 =c^2

?^2 + 4^2 = 10 ^2

?^2 + 16 = 100

Subtract 16 from each side

?^2 +16-16 = 100-16

?^2 = 84

Take the square root of each side

sqrt(?^2) = sqrt(84)

? = sqrt(4) sqrt(21)

? = 2 sqrt(21)

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3 years ago

(6+3)+21=6+(3+21) name each property of addition or multiplication

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4 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

3+4=19, 5+6=41, 1+3=?

Umnica [9.8K]

3+4 = 7. 19/7 = 2.71
5+6 = 11. 41/11 = 3.73
1+3 = 4. ?/4 = 4.75 <--- Following the pattern of +1.02 (2.71+1.02=3.73, therefore 3.73+1.02=4.75
We still need (?) ... ?/4=4.75 we can re-write this as 4(4.75)=?
4(4.75) = 19 (4x4=16, 4x0.75=3, 16+3=19)
Therefore, 1+3=19.

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Help I know is easy but I don't know ​

Help I know is easy but I don't know