All categories

3 years ago

Find the radius and circumference of a CD with a diameter of 4.75 inches. SOMEONE PLEASE HELP ME!!!!!

2 answers:

6 0

Here are your equations:
r = diameter/2
C = 2*pi*r

Now all you have to do is plug in the 4.75 for the diameter, and you have your answer!
If you don't have a calculator with "pi" on it, use 3.14159.

igomit [66]3 years ago

3 0

The radius is half of the diameter.
To find the circumference, use the equation C=2(pi)r or C=(pi)d.
r meaning radius. d meaning diameter.
(pi) is about 3.14 or 22/7

You might be interested in

How to solve complex number problem is x square + 2x +65=0

ad-work [718]

Ill teach you how to do some of these things so you won't be having to use this site

7 0

3 years ago

Kim drives x miles a week. Terry drives two times more than Kim and Jerry drives three times more than Terry. Which expression r

liberstina [14]

The expression for the amount that Jerry drives is 6x. This can be concluded since 2*3=6. And how to represent it should be described as 6x.

4 0

3 years ago

Sadie measured the floor of her storage unit, which is rectangular. It is 9 meters wide and 10 meters from one corner to the opp

Evgen [1.6K]

Answer:

lack

Step-by-step explanation:

4 0

3 years ago

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

f ' ' '(x) $= \dfrac{2}{(1+x)^3}$

f ' ' '(x) $= \dfrac{2}{(1+0)^3} = 2$

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001

Hence, the estimate of In(1.4) to the term is $\dfrac{0.4^5}{5}$ is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0

3 years ago

How do I find the absolute deviations? Will give Brainleist

goldenfox [79]

Answer:

Mean deviation is a statistical measure of the average deviation of values from the mean in a sample. It is calculated first by finding the average of the observations. The difference of each observation from the mean then is determined. The deviations then are averaged. This analysis is used to calculate how sporadic observations are from the mean.

Step-by-step explanation:

If there is an equation ill solve it for u

7 0

3 years ago