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3 years ago

Identify the diameter of the disc

Mathematics

1 answer:

Olegator [25]3 years ago

4 0

radius = (4*10^2 + 24^2)/8*10 =

(400 + 576)/80=

976/80 = 12.2

diameter = 12.2 x 2 = 24.4

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Is 7/13 greater than 1/2

blagie [28]

Answer:

Yes

Step-by-step explanation:

7/13*2=14/26

13/26=1/2

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3 years ago

What number has exactly 5 factors between 1 and 20

mariarad [96]

<span>The factors of 16 are 1, 2, 4, 8, and 16.</span>

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3 years ago

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What is the height of the building

Mrac [35]

I think the answer is 180m

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4 years ago

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I have calculus problems that I need help with.

aleksklad [387]

a. Note that $f(x)=x^ne^{-2x}$ is continuous for all $x$ . If $f(x)$ attains a maximum at $x=3$ , then $f'(3) = 0$ . Compute the derivative of $f$ .

$f'(x) = nx^{n-1} e^{-2x} - 2x^n e^{-2x}$

Evaluate this at $x=3$ and solve for $n$ .

$n\cdot3^{n-1} e^{-6} - 2\cdot3^n e^{-6} = 0$

$n\cdot3^{n-1} = 2\cdot3^n$

$\dfrac n2 = \dfrac{3^n}{3^{n-1}}$

$\dfrac n2 = 3 \implies \boxed{n=6}$

To ensure that a maximum is reached for this value of $n$ , we need to check the sign of the second derivative at this critical point.

$f(x) = x^6 e^{-2x} \\\\ \implies f'(x) = 6x^5 e^{-2x} - 2x^6 e^{-2x} \\\\ \implies f''(x) = 30x^4 e^{-2x} - 24x^5 e^{-2x} + 4x^6 e^{-2x} \\\\ \implies f''(3) = -\dfrac{486}{e^6} < 0$

The second derivative at $x=3$ is negative, which indicate the function is concave downward, which in turn means that $f(3)$ is indeed a (local) maximum.

b. When $n=4$ , we have derivatives

$f(x) = x^4 e^{-2x} \\\\ \implies f'(x) = 4x^3 e^{-2x} - 2x^4 e^{-2x} \\\\ \implies f''(x) = 12x^2 e^{-2x} - 16x^3e^{-2x} + 4x^4e^{-2x}$

Inflection points can occur where the second derivative vanishes.

$12x^2 e^{-2x} - 16x^3 e^{-2x} + 4x^4 e^{-2x} = 0$

$12x^2 - 16x^3 + 4x^4 = 0$

$4x^2 (3 - 4x + x^2) = 0$

$4x^2 (x - 3) (x - 1) = 0$

Then we have three possible inflection points when $x=0$ , $x=1$ , or $x=3$ .

To decide which are actually inflection points, check the sign of $f''$ in each of the intervals $(-\infty,0)$ , $(0, 1)$ , $(1, 3)$ , and $(3,\infty)$ . It's enough to check the sign of any test value of $x$ from each interval.

$x\in(-\infty,0) \implies x = -1 \implies f''(-1) = 32e^2 > 0$

$x\in(0,1) \implies x = \dfrac12 \implies f''\left(\dfrac12\right) = \dfrac5{43} > 0$

$x\in(1,3) \implies x = 2 \implies f''(2) = -\dfrac{16}{e^4} < 0$

$x\in(3,\infty) \implies x = 4 \implies f''(4) = \dfrac{192}{e^8} > 0$

The sign of $f''$ changes to either side of $x=1$ and $x=3$ , but not $x=0$ . This means only $\boxed{x=1}$ and $\boxed{x=3}$ are inflection points.

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1 year ago

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Givem the geometric sequence where à1=2 and the common ratio is 4, what is domain for N

Dmitry_Shevchenko [17]

The domain for N is All integers where n ≥ 1

<u>Solution:</u>

According to statement a1 = 2 and r = 4. This shows that r is greater than 1.

If r is greater than 1 than it includes integers greater than 1 or equal to 1. It does not include all the real numbers because real numbers include negative numbers also.

If starting value is 2, if we put n=0, then we get 2, but if we put a negative value than we would get a number which is not a part of our sequence.

Thus the domain of n is All integers where n greater than or equal to 1

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4 years ago