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olga nikolaevna [1]

3 years ago

14

A computer program in Shannon’s computer carries out a single mathematical operation in 1.5 × 10-6 seconds. How much time would

the program take to complete 2.5 × 103 mathematical operations?

1 answer:

kykrilka [37]3 years ago

3 0

1/1.5 * 10^-6 = 2500/x

x = 1.5 * 10^-6 * 2500

x = 0.00375 or 3.75 * 10^-3 second.

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8a to the power of 6 times a to the power of 4 over 4a to the power of -3

svetlana [45]

Answer:

(8a)⁻¹⁸

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Step-by-step explanation:

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((8a)^6)^-3

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3 years ago

Please help me with the below question.

VMariaS [17]

By letting

$y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}$

we get derivatives

$y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}$

$y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}$

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

$5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0$

Examine the lowest degree term $\left(x^{r-1}\right)$ , which gives rise to the indicial equation,

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with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients $c_k$ is

$(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}$

so that with r = 4/5, the coefficients are governed by

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$c_1 = -\dfrac{c_0}5 = -\dfrac15$

$c_2 = -\dfrac{c_1}{10} = \dfrac1{50}$

so that the first three terms of the solution are

$\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}$

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2 years ago

Please show all work with the answers please

kenny6666 [7]

Part A

The angle of elevation of the staire case is given by

$Angle\ of\ elevation=\tan^{-1}{ \frac{Vertical\ rise}{landing} } \\ \\ =\tan^{-1}{ \frac{14}{20} }\approx35^o$

Part B:

<span>If the landing of each step is changed to 17 inches and the elevation of each step to 21 inches, the new angle of elevation for the staircase is given by

</span><span> $Angle\ of\ elevation=\tan^{-1}{ \frac{Vertical\ rise}{landing} } \\ \\ =\tan^{-1}{ \frac{17}{21} }\approx39^o$ </span>

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3 years ago

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astra-53 [7]

√25/4=5/2 because 5/2 times 5/2 equal 25/4. Hope it help!

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3 years ago

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Paraphin [41]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers