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2 years ago

5) What decimal part of 40 is 8?

Mathematics

2 answers:

AfilCa [17]2 years ago

7 0

Answer: 0.2

Step-by-step explanation:

Nikolay [14]2 years ago

4 0

8/40 = 1/5
As a 1/5 decimal =0.2

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You are playing a solitaire game in which you are dealt three cards without replacement from a simplified deck of 10 cards (mark

g100num [7]

276 hands very simple im in 6th grade

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3 years ago

1. (5pts) Find the derivatives of the function using the definition of derivative.

andreyandreev [35.5K]

2.8.1

$f(x) = \dfrac4{\sqrt{3-x}}$

By definition of the derivative,

$f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$

We have

$f(x+h) = \dfrac4{\sqrt{3-(x+h)}}$

and

$f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}$

Combine these fractions into one with a common denominator:

$f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}$

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

$f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}$

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

$\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}$

3.1.1.

$f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3$

Differentiate one term at a time:

• power rule

$\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4$

$\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}$

$\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}$

The last two terms are constant, so their derivatives are both zero.

So you end up with

$f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}$

8 0

1 year ago

A survey showed that 23 out of every 100 students play a musical instrument. What percent of the students surveyed play a musica

Gnom [1K]

Answer:

23%

Step-by-step explanation:

7 0

3 years ago

8/7 divided by 5/8.

maksim [4K]

Answer:

64/35

Step-by-step explanation:

To divide fractions, multiply the first fraction by the second fraction's reciprocal.

The reciprocal of a fraction is just the top and bottom number switched.

So the reciprocal of 5/8 would be 8/5.

Now, you can multiply 8/7 by 8/5.

This gives you 64/35.

And that's it!

Please mark as Brainliest! :)

3 0

2 years ago

A researcher records the repair cost for 8 randomly selected washers. A sample mean of $60.46 and standard deviation of $18.36 a

Anuta_ua [19.1K]

Answer:

The critical value is T = 1.895.

The 90% confidence interval for the mean repair cost for the washers is between $48.159 and $72.761

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 8 - 1 = 6

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 6 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$ . So we have T = 1.895, which is the critical value.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.895\frac{18.36}{\sqrt{8}} = 12.301$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 60.46 - 12.301 = $48.159

The upper end of the interval is the sample mean added to M. So it is 60.46 + 12.301 = $72.761

The 90% confidence interval for the mean repair cost for the washers is between $48.159 and $72.761

8 0

2 years ago