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3 years ago

Anyone here in brainly good at math?

Mathematics

1 answer:

nalin [4]3 years ago

5 0

Answer:

3/16 / 5/6 = 9/40

7 / 1/6 = 42

2 1/6 / 1 1/3 = 1 13/20

2 1/7 / 3/28 = 20

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Of 120 owners of reptiles surveyed, 27 own snakes as pets. What percent of

maxonik [38]

Answer:

22.5%

Step-by-step explanation:

6 0

3 years ago

I need some help please

9966 [12]

Dilation:

The dilation factor is the length of image/preimage = 1.5

Thus
Q'R'/QR=1.5
Q'R'=1.5QR = 1.5 (5 cm) = 7.5 cm

(Edited)

8 0

3 years ago

Read 2 more answers

Given: BC = AD, AD = 17 DC = 14, m∠1=m∠2 Find: AB, BC

Mariulka [41]

Answer:

AB = 14 and BC = 17

Step-by-step explanation:

Congruence is a property that shows an equal relation between sides or angles of a given figure or two figures.

From the question, the given conditions are;

BC = AD, AD = 17, DC = 14, and m< 1 = m<2 .

i. Considering the property of congruence of sides,

BC = 17 (∵ BC = AD and AD has been given to equal 17)

i.e BC = AD = 17

ii. Considering the property of congruent angles and sides,

AB = 14 (∵ m<1 = m<2 and DC is 14)

Therefore, AB = 14 and BC = 17.

6 0

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

6 0

3 years ago

Decribe the counter clockwise rotation that maps angles

Llana [10]

I don’t know if this is the answer but.. An angle generated by one complete counterclockwise rotation measures 360° or 21 radians. 360° An angle generated by one complete clockwise rotation measures -360° or -21 radians.

5 0

3 years ago