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2 years ago

What number is one thousand more than 3,449?

Mathematics

1 answer:

yKpoI14uk [10]2 years ago

3 0

Answer:

4,449

Step-by-step explanation:

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Answer:

equilateral triangle 3 3 perpendicular bisectors of the sides

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regular octagon 8 4 (perpendicular bisectors and 4 lines joining pairs of opposite vertices)

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Step-by-step explanation:

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2 years ago

divide 24 fish into five groups. What is the remainder of this division? Hint: Group the fish into sets of 5, and count how many

vlada-n [284]

Answer:

4.4

Step-by-step explanation:

The answer is 4.4 because if you know that 5*4 equal 20 so 4 can go each of the 5 groups so 24-20 leaves you with 4 so your answer is 4 remainder 4

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Jason and David went hiking at different times. Jason is currently 54 feet above sea level while David is starting at 11 feet be

tatuchka [14]

Since jason is currently at 54 feet above sea level and David is 11 feet below sea level the add 11 plus 54 which would give you 65 feet is there distance of elevations

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3 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

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

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

4 0

2 years ago