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3 years ago

I need help please for my test

Mathematics

1 answer:

Umnica [9.8K]3 years ago

7 0

original figure (-7,-6)

final figure (-7,2)

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Morgarella [4.7K]

Answer:

$f'(x)=-\frac{2}{x^\frac{3}{2}}$

Step-by-step explanation:

So we have the function:

$f(x)=\frac{4}{\sqrt x}$

And we want to find the derivative using the limit process.

The definition of a derivative as a limit is:

$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Therefore, our derivative would be:

$\lim_{h \to 0}\frac{\frac{4}{\sqrt{x+h}}-\frac{4}{\sqrt x}}{h}$

First of all, let's factor out a 4 from the numerator and place it in front of our limit:

$=\lim_{h \to 0}\frac{4(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x})}{h}$

Place the 4 in front:

$=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}$

Now, let's multiply everything by (√(x+h)(√(x))) to get rid of the fractions in the denominator. Therefore:

$=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}(\frac{\sqrt{x+h}\sqrt x}{\sqrt{x+h}\sqrt x})$

Distribute:

$=4\lim_{h \to 0}\frac{({\sqrt{x+h}\sqrt x})\frac{1}{\sqrt{x+h}}-(\sqrt{x+h}\sqrt x)\frac{1}{\sqrt x}}{h({\sqrt{x+h}\sqrt x})}$

Simplify: For the first term on the left, the √(x+h) cancels. For the term on the right, the (√(x)) cancel. Thus:

$=4 \lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }$

Now, multiply both sides by the conjugate of the numerator. In other words, multiply by (√x + √(x+h)). Thus:

$= 4\lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }(\frac{\sqrt x +\sqrt{x+h})}{\sqrt x +\sqrt{x+h})}$

The numerator will use the difference of two squares. Thus:

$=4 \lim_{h \to 0} \frac{x-(x+h)}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Simplify the numerator:

$=4 \lim_{h \to 0} \frac{x-x-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}\\=4 \lim_{h \to 0} \frac{-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Both the numerator and denominator have a h. Cancel them:

$=4 \lim_{h \to 0} \frac{-1}{(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Now, substitute 0 for h. So:

$=4 ( \frac{-1}{(\sqrt{x+0}\sqrt x)(\sqrt x+\sqrt{x+0})})$

Simplify:

$=4( \frac{-1}{(\sqrt{x}\sqrt x)(\sqrt x+\sqrt{x})})$

(√x)(√x) is just x. (√x)+(√x) is just 2(√x). Therefore:

$=4( \frac{-1}{(x)(2\sqrt{x})})$

Multiply across:

$= \frac{-4}{(2x\sqrt{x})}$

Reduce. Change √x to x^(1/2). So:

$=-\frac{2}{x(x^{\frac{1}{2}})}$

Add the exponents:

$=-\frac{2}{x^\frac{3}{2}}$

And we're done!

$f(x)=\frac{4}{\sqrt x}\\f'(x)=-\frac{2}{x^\frac{3}{2}}$

5 0

3 years ago

Car X weighs 136 pounds more than car Z. Car Y weighs 117 pounds more than car Z. The total weight of all three cars is 9439 pou

Aleksandr [31]

Let x, y and z denote the weighs of car X, car Y and car Z, respectively.

We know that car X weighs 136 more than car Z, this can be express by the equation:

$x=z+136$

We also know that Y weighs 117 pounds more than car Z, this can be express as:

$y=z+117$

Finally, we know that the total weight of all the cars is 9439, then we have:

$x+y+z=9439$

Hence, we have the system of the equations:

$\begin{gathered} x=z+136 \\ y=z+117 \\ z+y+z=9439 \end{gathered}$

To solve the system we can plug the values of x and y, given in the first two equations, in the last equation; then we have:

$\begin{gathered} z+136+z+117+z=9439 \\ 3z=9439-136-117 \\ 3z=9186 \\ z=\frac{9186}{3} \\ z=3062 \end{gathered}$

Now that we have the value of z we plug it in the first two equations to find x and y:

$\begin{gathered} x=3062+136=3198 \\ y=3062+117=3179 \end{gathered}$

Therefore, car X weighs 3198 pound, car Y weighs 3179 pounds and car Z weighs 3062 pounds.

4 0

1 year ago

The function C(x) = 500(1-0.011)* models the amount of antibiotic, in milligrams, in the body x minutes after

kirill115 [55]

Answer:

Step-by-step explanation:

Ur mum

4 0

2 years ago

Suppose a geyser has a mean time between eruptions of 72 minutes. Let the interval of time between the eruptions be normally dis

nikitadnepr [17]

Answer:

(a) The probability that a randomly selected time interval between eruptions is longer than 82 minutes is 0.3336.

(b) The probability that a random sample of 13-time intervals between eruptions has a mean longer than 82 minutes is 0.0582.

(c) The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 minutes is 0.0055.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) The population mean must be more than 72, since the probability is so low.

Step-by-step explanation:

We are given that a geyser has a mean time between eruptions of 72 minutes.

Also, the interval of time between the eruptions be normally distributed with a standard deviation of 23 minutes.

(a) Let X = the interval of time between the eruptions

So, X ~ N( $\mu=72, \sigma^{2} =23^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 72 minutes

$\sigma$ = standard deviation = 23 minutes

Now, the probability that a randomly selected time interval between eruptions is longer than 82 minutes is given by = P(X > 82 min)

P(X > 82 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{82-72}{23}$ ) = P(Z > 0.43) = 1 - P(Z $\leq$ 0.43)

= 1 - 0.6664 = 0.3336

The above probability is calculated by looking at the value of x = 0.43 in the z table which has an area of 0.6664.

(b) Let $\bar X$ = sample mean time between the eruptions

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time = 72 minutes

$\sigma$ = standard deviation = 23 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 minutes is given by = P( $\bar X$ > 82 min)

P( $\bar X$ > 82 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{82-72}{\frac{23}{\sqrt{13} } }$ ) = P(Z > 1.57) = 1 - P(Z $\leq$ 1.57)

= 1 - 0.9418 = 0.0582

The above probability is calculated by looking at the value of x = 1.57 in the z table which has an area of 0.9418.

(c) Let $\bar X$ = sample mean time between the eruptions

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time = 72 minutes

$\sigma$ = standard deviation = 23 minutes

n = sample of time intervals = 34

Now, the probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 minutes is given by = P( $\bar X$ > 82 min)

P( $\bar X$ > 82 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{82-72}{\frac{23}{\sqrt{34} } }$ ) = P(Z > 2.54) = 1 - P(Z $\leq$ 2.54)

= 1 - 0.9945 = 0.0055

The above probability is calculated by looking at the value of x = 2.54 in the z table which has an area of 0.9945.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) If a random sample of 34-time intervals between eruptions has a mean longer than 82 minutes, then we conclude that the population mean must be more than 72, since the probability is so low.

6 0

3 years ago

Please help me, I will mark brainlieat

Minchanka [31]

Answer:

c or 1,500

Step-by-step explanation:

hope this helps lol

4 0

3 years ago

Read 2 more answers