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3 years ago

Find the common ratio for the following sequence. 0.1, 0.01, 0.001,

Mathematics

1 answer:

nika2105 [10]3 years ago

6 0

Answer:

The common ratio is 0.10

Step-by-step explanation:

0.1 x 0.1 = 0.01

0.01 x 0.1 = 0.001

common ratio is the ratio of a term to the previous term.

0.1 is the previous term of 0.01

common ratio = 0.01/0.1 = 0.10

common ratio = 0.001/0.01 = 0.10

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Chapter 12 F

marshall27 [118]

Answer:

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Step-by-step explanation:

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3 years ago

Maria measured a centipede that was 1 1/12 inches long. Jerome measured a centipede that was 11/12 of an inch long. How much lon

4vir4ik [10]

Answer:

1/6

Step-by-step explanation:

The centipede that Maria measured was 1 1/12 inches long. If you convert this to an improper fraction it is equal to 13/12 inches. Now to find how much longer Maria's centipede was than Jeromes, you just take 11/12 and minus it from 13/12 which is equal to 2/12 of an inch, which then simplifies to 1/6 of an inch.

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3 years ago

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Rudik [331]

Answer:

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2 years ago

You stand a known distance from the base of the tree, measure the angle of elevation the top of the tree to be 15â—¦ , and then

gogolik [260]

Answer:

The maximum possible error of in measurement of the angle is $d\theta_1 =(14.36p)^o$

Step-by-step explanation:

From the question we are told that

The angle of elevation is $\theta_1 = 15 ^o = \frac{\pi}{12}$

The height of the tree is h

The distance from the base is D

h is mathematically represented as

$h = D tan \theta$ Note : this evaluated using SOHCAHTOA i,e

$tan\theta = \frac{h}{D}$

Generally for small angles the series approximation of $tan \theta \ is$

$tan \theta = \theta + \frac{\theta ^3 }{3}$

So given that $\theta = 15 \ which \ is \ small$

$h = D (\theta + \frac{\theta^3}{3} )$

$dh = D (1 + \theta^2) d\theta$

=> $\frac{dh}{h} = \frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta$

Now from the question the relative error of height should be at most

$\pm p$ %

=> $\frac{dh}{h} = \pm p$

=> $\frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta = \pm p$

=> $d\theta = \pm \frac{\theta + \frac{\theta^3}{3} }{1+ \theta ^2} * \ p$

So for $\theta_1$

$d\theta_1 = \pm \frac{\theta_1 + \frac{\theta^3_1 }{3} }{1+ \theta_1 ^2} * \ p$

substituting values

$d [\frac{\pi}{12} ] = \pm \frac{[\frac{\pi}{12} ] + \frac{[\frac{\pi}{12} ]^3 }{3} }{1+ [\frac{\pi}{12} ] ^2} * \ p$

=> $d\theta_1 = 0.25 p$

Converting to degree

$d\theta_1 = (0.25* 57.29) p$

$d\theta_1 =(14.36p)^o$

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3 years ago

I need help with these...

DaniilM [7]

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3 years ago