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djverab [1.8K]
2 years ago
14

The point P(7, −4) lies on the curve y = 4/(6 − x). (a) If Q is the point (x, 4/(6 − x)), use your calculator to find the slope

mPQ of the secant line PQ (correct to six decimal places) for the following values of x. (i) 6.9 mPQ = (ii) 6.99 mPQ = (iii) 6.999 mPQ = (iv) 6.9999 mPQ = (v) 7.1 mPQ = (vi) 7.01 mPQ = (vii) 7.001 mPQ = (viii) 7.0001 mPQ = (b) Using the results of part (a), guess the value of the slope m of the tangent line to the curve at P(7, −4). m = (c) Using the slope from part (b), find an equation of the tangent line to the curve at P(7, −4).
Mathematics
1 answer:
Luda [366]2 years ago
6 0
Hopefully you have a Ti-84 calculator because that's what I'm going to use.

First, input the equation on the "y =" button and graph the equation.

Then, hit "2nd, trace", or "calc". Choose the 6th option, "dy/dx". This gives you the derivative, or slope of the secant line, at any given x value.

If you just start typing numbers and hit enter, it'll find the derivative, so for part a letters i-viii, just input those numbers and press enter.

From the looks of it, the slope approaches 4 (as you from the x-values in i-iv and v-viii, the slopes approach 4). You can also check this by just inputting the x-value 7 in dy/dx or taking the derivative of the equation and plugging in 7. Hope this helps!! If you liked this answer please rate it as brainliest!!!
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Simplify the answer
Andru [333]

Answer:

- 20 + 8i

Step-by-step explanation:

Noting that i²  = - 1

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8 0
3 years ago
Assume the hold time of callers to a cable company is normally distributed with a mean of 5.5 minutes and a standard deviation o
Ierofanga [76]

Answer:

The percent of callers are 37.21 who are on hold.

Step-by-step explanation:

Given:

A normally distributed data.

Mean of the data, \mu = 5.5 mins

Standard deviation, \sigma = 0.4 mins

We have to find the callers percentage who are on hold between 5.4 and 5.8 mins.

Lets find z-score on each raw score.

⇒ z_1=\frac{x_1-\mu}{\sigma}   ...raw score,x_1 = 5.4

⇒ Plugging the values.

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⇒ z_1=-0.25  

For raw score 5.5 the z score is.

⇒ z_2=\frac{5.8-5.5}{0.4}  

⇒ z_2=0.75

Now we have to look upon the values from Z score table and arrange them in probability terms then convert it into percentages.

We have to work with P(5.4<z<5.8).

⇒ P(5.4

⇒ P(-0.25

⇒ P(z

⇒ z(1.5)=0.7734 and z(-0.25)=0.4013.<em>..from z -score table.</em>

⇒ 0.7734-0.4013

⇒ 0.3721

To find the percentage we have to multiply with 100.

⇒ 0.3721\times 100

⇒ 37.21 %

The percent of callers who are on hold between 5.4 minutes to 5.8 minutes is 37.21

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natima [27]

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Step-by-step explanation:

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Arturiano [62]
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jok3333 [9.3K]
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