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3 years ago

Which of the following rules could represent the function shown in the table?

Mathematics

1 answer:

MariettaO [177]3 years ago

6 0

Answer:

f(x) = 2x+1

Step-by-step explanation:

Equation in slope intercept form y = mx + b where m = slope b = y-intercept

To find slope ,

m = (y2-y1)/(x2 - x1)

m = (3 - 1) / (1 - 0)

m = 32/1

m = 2

y-intercept b = 1

In this case y = 2x + 1 or f(x) = 2x+1

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Complete the proof of ∆JKL ≅ ∆LMN, by providing the reason for each of the statements below. Reasons MAY be used more than once.

aksik [14]

Answer:

See below

Step-by-step explanation:

1. JK ∥ LM, KL ∥ MN Given

2. ∠KJL ≅ ∠MLN; ∠KLJ ≅ ∠MNL c. Corresponding Angles

3. JK ≅ LM Given

4. ∆JKL ≅ ∆LMN e. AAS

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3 years ago

KL is congruent toMN and your answer.

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3 years ago

A cable company charges $70 per month for cable service to existing customers

Pachacha [2.7K]

Answer:

70x=y

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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

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

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

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3 years ago

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adoni [48]

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4 years ago