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3 years ago

What is the 20th term in the following arithmatic sequence 1/2, 1, 3/2, 2...

1 answer:

4 0

The 20th term in the following arithmetic sequence1/2, 1, 3/2, 2... is 10 because;

the number are divided by 2
and the 20th number would be 20/2 which is equal to 10

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1.2cm represents 24 km. it is approximately 5 cm from knight rock to storm bay on the map. what is the actual distance in kilome

Kipish [7]

Answer:

about 120

Step-by-step explanation:

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3 years ago

The force exerted by an electric charge at the origin on a charged particle at a point (x, y, z) with position vector r = x, y,

suter [353]

Answer:

$\frac{25k}{34}[\frac{1}{4}-\frac{1}{50^{1/2}}] \\$

Step-by-step explanation:

The straight line path between point (a,b,c) and (l,m,n) is parametric by the expression

$r(t)=(1-t)(a,b,c) +t(l,m,n)\\$

since we are giving point (4,0,0) and (4,3,4), the parametric equation is giving below

$r(t)=(1-t)(4,0,0) +t(4,3,4)\\$

using the dot product system of multiplication, we have

$r(t)=(4,3t,4t) \\$

t is between 0,1.

Next we define the line integral for work done which is express as

$\int\limits^a_b {F.} \,dr\\$

First we define the general expression for the force

$f(x,y,z)=\frac{K(x,y,z)}{(x^{2}+y^{2}+z^{2})^{3/2}} \\$

If we substitute our parametric equation we arrive at

$F(r(t))=\frac{K(4,3t,4t)}{(4^{2}+(3t)^{2}+(4t)^{2})^{3/2}}\\F(r(t))=\frac{K(4,3t,4t)}{(16+34t^{2})^{3/2}}\\$

also we need to find the expression for <em>dr</em>

$r(t)=(4,3t,4t)\\dr=(0,3,4)dt\\$

Now we substitute into the integral expression

$\int\limits^1_0 {\frac{k(4,3t,4t)}{(16+34t^{2})^{3/2}}} \, .(0,3,4)dt$

using dot product we arrive at

$\int\limits^1_0 {\frac{25kt}{{(16+34t^{2})^{3/2}} } \,$

let make a simple substitution so we can simplify the integral,

let assume $u=16+34t^{2}\\$

$\frac{du}{dt}=68t\\ dt=\frac{du}{68t} \\$

and changing setting the new upper and lower limit, we have

$\frac{25k}{68} \int\limits^a_b {\frac{1}{u\frac{3}{2} } } \, du\\a=50\\b=16\\$

by simple integral we arrive at

$-\frac{25k}{34}[\frac{1}{u^{1/2}}] ^{50}_{16} \\\frac{25k}{34}[\frac{1}{4}-\frac{1}{50^{1/2}}] \\$

Hence the workdone is

$\frac{25k}{34}[\frac{1}{4}-\frac{1}{50^{1/2}}] \\$

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3 years ago

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3 years ago

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3 years ago

If the distribution of the population has a nonzero standard deviation and mean 10 then the probability that an individual data

RoseWind [281]

Answer:

FALSE

Step-by-step explanation:

The central limit theorem tell us that for a random sample of size n, the average of the sample will be approximately normally distributed with mean $\mu$ and variance $\sigma^2/n$ , where $\mu$ and $\sigma^2$ are the mean a variance respectively of the sampling distribution. This always that the mean and the variance are both finite and the sample size n is greater than 30. We could use the central limit theorem, but in this case does not help us because we are considering 25 randomly selected data values, besides, we do not know the distribution of the original population.

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3 years ago