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3 years ago

Which is the graph of y=2/(x+1)-6

Mathematics

2 answers:

Mariana [72]3 years ago

8 0

Answer:

A on EDGE2020

krek1111 [17]3 years ago

4 0

Answer:

I have linked the graph in an image

Step-by-step explanation:

1. Simplify y=2/(x+1)-6

2. Subtract the numbers 1-6=-5

3. = 2/x-5

4. Graph your results

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1. Sand falling from a chute forms a conical pile whose altitude is always equal to 4/3 the radius of the base. (a) How fast is

solong [7]

Answer:

The volume is increasing at a rate of 16286 in³/min.

Step-by-step explanation:

a) The volume of a cone is given by:

$V = \frac{1}{3}h\pi r^{2}$

Where:

r: is the radius

h: is the height

The rate of change of the volume can be calculated by using the chain rule:

$\frac{dV}{dt} = \frac{\pi}{3}[\frac{dh}{dt}r^{2} + h\frac{d(r^{2})}{dt}]$

Since h = 4/3 r we have:

$\frac{dV}{dt} = \frac{\pi}{3}[\frac{d(\frac{4r}{3})}{dt}r^{2} + \frac{4r}{3}\frac{d(r^{2})}{dt}]$

$\frac{dV}{dt} = \frac{4\pi}{9}[\frac{dr}{dt}r^{2} + r\frac{d(r^{2})}{dt}]$

$\frac{dV}{dt} = \frac{4\pi}{9}[\frac{dr}{dt}r^{2} + 2r^{2}\frac{dr}{dt}]$ (1)

With:

$\frac{dr}{dt}$ = 3 in/min

$r = 3 ft*\frac{12 in}{1 ft} = 36 in$

And by entering the above values into equation (1) we have:

$\frac{dV}{dt} = \frac{4\pi}{9}[(3 in/min)*(36 in)^{2} + 2*(36 in)^{2}*3 in/min]$

$\frac{dV}{dt} = 16286 in^{3}/min$

Therefore, the volume is increasing at a rate of 16286 in³/min.

I hope it helps you!

3 0

3 years ago

Use the product, quotient, and power rules of logarithms to rewrite the expression as a single logarithm. Assume that all variab

Firlakuza [10]

Answer:

log x² $y^{4}$

Step-by-step explanation:

Using the rules of logarithms

log x + log y ⇔ log xy

log $x^{n}$ ⇔ n log x

Given

2logx + 4logy

= logx² + log $y^{4}$

= log x² $y^{4}$

3 0

3 years ago

A large rectangle is made by joining three identical small rectangles as shown.

vampirchik [111]

Answer: 35 cm

Step-by-step explanation:

As shown in the image attached, the A large rectangle is made by joining three identical small rectangles,

The width of one small rectangle is x cm and the length of one small rectangle is 2x cm. Therefore the perimeter of the small rectangle is given as:

2(length + width) = Perimeter

2(2x + x) = 21

2(3x) = 21

6x = 21

x = 21/6 = 3.5 cm

x = 3.5 cm

From the image attached, the width of the large rectangle is 2x (x + x) and the length is 3x (2x + x). Therefore, the perimeter of the large rectangle is:

2(length + width) = Perimeter

2(3x + 2x) = Perimeter

Perimeter = 2(5x)

Perimeter = 10x

Perimeter = 10(3.5)

Perimeter = 35 cm

Step-by-step explanation:

3 0

2 years ago

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in he

UkoKoshka [18]

Answer:

a) The minimum sample size is 601.

b) The minimum sample size is 2401.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

For this problem, we have that:

We dont know the true proportion, so we use $\pi = 0.5$ , which is when we are are going to need the largest sample size.

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.04\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.04}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2$

$n = 600.25$

Rounding up

The minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}$