A cylinder has a radius of 5 cm and a height of 8 cm. How many cubic meters of fluid can fill this cylinder

How many permutations<br> of the letters ANSWER<br> end in a vowel?

GREYUIT [131]

Update the question so it's on-topic for Mathematics Stack Exchange. Closed 4 years ago. How many words can be made by using the letters of the word ENGLAND such that words begin and end with a vowel? I couldn't understand how come the answer to this question is 120

6 0

3 years ago

A right triangle has sides of lengths 18 , 24 , and 30 units. What is the area of the triangle? Draw the shape on a grid to h

Elden [556K]

Answer:216

Step-by-step explanation:The formula to find the area of a triangle is Length times Base divided by 2. The length of the triangle could be 18 or 24, but that doesn’t matter. The base could also be 18 or 24, but that also doesn’t matter, because the hypotenuse (the longest part of a right triangle, in this case being 30), is not a part of the formula. 18 times 24 is 432, and 432 divided by 2 is 216. So the area is 216

7 0

3 years ago

The height 'h' (in feet) of a ball in a baseball game can be modeled by h = -16t² + 28t + 8 , where 't' is the time (in seconds)

lidiya [134]

Answer:

a) No. t < 0 is not part of the useful domain of the function

b) 2.0 seconds

Step-by-step explanation:

a) A graph of the function is shown below. It shows t-intercepts at t=-0.25 and t=2.0. We presume that t is measured forward from some event such as the ball being thrown or hit. The model's predicted ball location has no meaning prior to that event, when values of t are negative.

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b) It is convenient to use a graphing calculator to find the t-intercepts. Or, the equation can be solved for h=0 any of several ways algebraically. One is by factoring.

h = 0 = -16t² +28t +8 . . . . . . . . . . . . the ball hits the ground when h = 0

0 = -4(4t² -7t -2) = -4(4t +1)(t -2)

This has t-intercepts where the factors are zero, at t=-1/4 and t=2.

The ball will hit the ground after 2 seconds.

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