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3 years ago

Simplify to create an equivalent expression 19−6(−k−2)

Mathematics

1 answer:

lesya [120]3 years ago

5 0

Answer:

boots and cuts

Step-by-step explanation:

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Find the area of the region that lies inside the first curve and outside the second curve.

marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = $\dfrac{1}{2}$

$\theta = -\dfrac{\pi}{3}, \ \ \dfrac{\pi}{3}$

Now, the area of the region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \ \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ 5^2 d \theta$

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \ \theta d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ d \theta$

$A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} \dfrac{cos \ 2 \theta +1}{2} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}$

$A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} {cos \ 2 \theta +1} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}} \ \ - \dfrac{25}{2} \begin {bmatrix} \dfrac{2 \pi}{3} \end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3}) \end {bmatrix} - \dfrac{25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} + \dfrac{\pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}$

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx

7 0

3 years ago

Matthew's mother asked him to go to the store for her. To get to the store, he walked seven and a half city blocks. He caught th

Ratling [72]

Answer:

matthew travel 22 blocks i think

4 0

3 years ago

The amount of protein that an individual must consume is different for every person. There are solid theoretical ideas that sugg

amid [387]

Answer:

The proportion of the population that have a protein requirement less than 0.60 g P • kg-1 • d-1 is 0.239, that is, 239 persons for every 1000, or simply 23.9% of them.

$\\ 0.239 =\frac{239}{1000}\;or\;23.9\%$

Step-by-step explanation:

From the question, we have the following information:

The distribution for protein requirement is normally distributed.
The population mean for protein requirement for adults is $\\ \mu= 0.65 gP*kg^{-1}*d^{-1}$
The population standard deviation is $\\ \sigma =0.07 gP*kg^{-1}*d^{-1}$

We have here that protein requirements in adults is normally distributed with defined parameters. The question is about the proportion of the population that has a requirement less than $\\ x = 0.60 gP*kg^{-1}*d^{-1}$ .

For answering this, we need to calculate a z-score to obtain the probability of the value x in this distribution using a standard normal table available on the Internet or on any statistics book.

<h3>z-score</h3>

A z-score is expressed as

$\\ z = \frac{x - \mu}{\sigma}$

For the given parameters, we have:

$\\ z = \frac{0.60 - 0.65}{0.07}$

$\\ z = -0.7142857$

<h3>Determining the probability</h3>

With this value for z at hand, we need to consult a standard normal table to determine what the probability of this value is.

The value for z = -0.7142857 is telling us that the requirement for protein is below the population mean (negative sign indicates this). However, most standard normal tables give a probability that a statistic is less than z and for values greater than the mean (in other words, positive values). To overcome this, we need to take the complement of the probability given for z-score z = 0.7142857, that is, subtract from 1 this probability, which is possible because the normal distribution is symmetrical.

Tables have values for z with two decimal places, then, for z = 0.7142857, we need to rewrite it as z = 0.71. For this value, the standard normal table gives a value of P(z<0.71) = 0.76115.

Therefore, the cumulative probability for values less than x = 0.60 which corresponds to a z-score = -0.7142857 is approximately:

$\\ P(x$