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3 years ago

Solve the inequality IxI < 5

Mathematics

1 answer:

Elodia [21]3 years ago

3 0

Answer:

the correct answer is the second one

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IgorLugansk [536]

Answer:

9.53 cm2

Step-by-step explanation:

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

6 0

3 years ago

Simplify: (6)(93). Write your answer using an exponent.

strojnjashka [21]

Answer:

toooook teaehes

Step-by-step explanation:

sedmee endm the worlds nssssssssssssss

8 0

2 years ago

MAKE YOU THE BRAINLIEST EASY QUESTION Show the formula for finding the area of a triangle. Then find the area of a right triangl

lesya692 [45]

Answer:

Formula to find the area of a triangle is 1/2(bxh). The area of the right triangle is 28.

Step-by-step explanation:

To solve this problem you would plug in the base and height into the formula and follow the order of operations (P.E.M.D.A.S). So the equation would then read 1/2(5x11.2). Once you solve inside the parenthesis, the equation then reads, 1/2(56). Then you solve what half of 56 is, which is 28.

I hoped this helped you!

6 0

3 years ago

Read 2 more answers

In 2011, a U.S. Census report determined that 71% of college students work. A researcher thinks this percentage has changed sinc

coldgirl [10]

Answer:

Note: The full question is attached as picture below

a) Hо : p = 0.71

Ha : p ≠ 0.71

p = x / n

p = 91/110

p = 0.83.

1 - Pо = 1 - 0.71 = 0.29.

b) Test statistic = z

= p - Pо / [√Pо * (1 - Pо ) / n]

= 0.83 - 0.71 / [√(0.71 * 0.29) / 110]

= 0.12 / 0.043265

= 2.77360453

Test statistic = 2.77

c) P-value

P(z > 2.77) = 2 * [1 - P(z < 2.77)] = 2 * 0.0028

P-value = 0.0056

∝ = 0.01

P-value < ∝

Reject the null hypothesis. There is sufficient evidence to support the researchers claim at the 1% significance level.

6 0

3 years ago

Solve the inequality IxI < 5​

Solve the inequality IxI < 5