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3 years ago

If x = -2, which inequality is true?

Mathematics

2 answers:

Anettt [7]3 years ago

4 0

It’s J, 3 - X would equal 3 - (-2) which turns into 3+2 which equals 5 but 5 is not greater than 10

kaheart [24]3 years ago

4 0

Answer:

3 - x <10

Step-by-step explanation:

plug in -2 as x in each equation and see if its true

2(-2) + 6 < -1 -- false

2(-2) - 3 > 6 false

-5 + 3(-2) > 1 false

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A plant inspector takes a random sample of six month old seedlings from a nursery and measures their heights to the nearest mm .

morpeh [17]

Answer:

Step-by-step explanation:

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3 years ago

V over 4 minus 3 equals 5, what does v equal?

Lyrx [107]

Answer:

Step-by-step explanation:

v/4 -3=5

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v/4=8

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3 years ago

Integrate the following

enot [183]

I suppose you mean to have the entire numerator under the square root?

$\displaystyle\int_2^4\frac{\sqrt{x^2-4}}{x^2}\,\mathrm dx$

We can use a trigonometric substitution to start:

$x=2\sec t\implies\mathrm dx=2\sec t\tan t\,\mathrm dt$

Then for $x=2$ , $t=\sec^{-1}1=0$ ; for $x=4$ , $t=\sec^{-1}2=\frac\pi3$ . So the integral is equivalent to

$\displaystyle\int_0^{\pi/3}\frac{\sqrt{(2\sec t)^2-4}}{(2\sec t)^2}(2\sec t\tan t)\,\mathrm dt=\int_0^{\pi/3}\frac{\tan^2t}{\sec t}\,\mathrm dt$

We can write

$\dfrac{\tan^2t}{\sec t}=\dfrac{\frac{\sin^2t}{\cos^2t}}{\frac1{\cos t}}=\dfrac{\sin^2t}{\cos t}=\dfrac{1-\cos^2t}{\cos t}=\sec t-\cos t$

so the integral becomes

$\displaystyle\int_0^{\pi/3}(\sec t-\cos t)\,\mathrm dt=\boxed{\ln(2+\sqrt3)-\frac{\sqrt3}2}$

7 0

3 years ago

A piece of wire of length 5050 is cut, and the resulting two pieces are formed to make a circle and a square. Where should the

algol [13]

Answer:

x should be cut at 2221.5 to minimize the total combined area, and at 5050 to maximize it.

Step-by-step explanation:

Let x be the length of wire that is cut to form a circle within the 5050 wire, so 5050 - x would be the length to form a square.

A circle with perimeter of x would have a radius of x/(2π), and its area would be

$A_c = \pir^2 = \pi (\frac{x}{2pi})^2 = \frac{x^2}{4\pi}$

A square with perimeter of 5050 - x would have side length of (5050 - x)/4, and its area would be

$A_s = (\frac{5050 - x}{4})^2 = \frac{(5050 - x)^2}{16}$

The total combined area of the square and circles is

$\sum A = A_c + A_s = \frac{x^2}{4\pi} + \frac{(5050 - x)^2}{16}$

To find the maximum and minimum of this, we just take the 1st derivative, and set it to 0

$A' = \frac{2x}{4\pi} + \frac{-2(5050-x)}{16} = 0$

$\frac{x}{2\pi} - \frac{5050 - x}{8} = 0$

Multiple both sides by 8π and we have

$4x - 5050\pi + x\pi = 0$

$x(4 + \pi) = 5050\pi$

$x = \frac{5050\pi}{4 + \pi} = 2221.5$

At x = 2221.5:

$A = \frac{x^2}{4\pi} + \frac{(5050 - x)^2}{16}$ = 392720 + 500026 = 892746 [/tex]

At x = 0, $A = 5050^2/16 = 1593906$

At x = 5050, $A = \frac{5050^2}{4\pi} = 2029424$

As 892746 < 1593906 < 2029424, x should be cut at 2221.5 to minimize the total combined area, and at 5050 to maximize it.

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3 years ago

The graph of g(x)

lara31 [8.8K]

Answer:

If f(x)=x^2 and you translate it's graph 8 units to the left and 2 units up, g(x)=(x^2+8)+2

8 0

4 years ago

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