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3 years ago

9 m 3 m 3 m What is the area, in square meters, of the shaded region in the diagram?

Mathematics

1 answer:

NeTakaya3 years ago

3 0

Answer:

Notice that the side length of the square is 38 m.

area of square = (side length)2 = (38 m)2 = 1444 m2

Notice that the diameter of the circle is 38 m .

And its radius = diameter / 2 = 38m / 2 = 19 m

area of circle = pi * radius2 = pi * (19m)2 = pi * 361 m2 ≈ 1133.54 m2

So......

area of shaded region = area of square - area of circle

area of shaded region ≈ 1444 m2 - 1133.54 m2

area of shaded region ≈ 310.46 m2

Step-by-step explanation:

Notice that the side length of the square is 38 m.

area of square = (side length)2 = (38 m)2 = 1444 m2

Notice that the diameter of the circle is 38 m .

And its radius = diameter / 2 = 38m / 2 = 19 m

area of circle = pi * radius2 = pi * (19m)2 = pi * 361 m2 ≈ 1133.54 m2

So......

area of shaded region = area of square - area of circle

area of shaded region ≈ 1444 m2 - 1133.54 m2

area of shaded region ≈ 310.46 m2

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3 years ago

Read 2 more answers

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.

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3 years ago

In a word processing document or on a separate piece of paper, use the guide to construct a two-column proof proving AC > EF

VARVARA [1.3K]

Answer:

Complete proof is written below.

Facts and explanation about the segments shown in question :

As BC = EF is a given statement in the question
AB + BC = AC because the definition of betweenness gives us a clear idea that if a point B is between points A and C, then the length of AB and the length of BC is equal to the length of AC. Also according to Segment addition postulate, AB + BC = AC. For example, if AB = 5 and BC= 7 then AC = AB + BC → AC = 12
AC > BC because the Parts Theorem (Segments) mentions that if B is a point on AC between A and C, then AC > BC and AC>AB. So, if we observe the question figure, we can realize that point B lies on the segment AC between points A and C.
AC > EF because BC is equal to EF and if AC>BC, then it must be true that the length of AC must greater than the length segment EF.

Below is the complete proof of the observation given in the question:

STATEMENT REASON 

___________________________________________________

1. BC = EF 1. Given

2. AB + BC = AC 2. Betweenness

3. AC > BC 3. Def. of segment inequality

4. AC > EF 4. Def. of congruent segments

Keywords: statement, length, reason, proof

Learn more about proof from brainly.com/question/13005321

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3 years ago

A car is purchased for $30,000. The value of the car depreciates annually so that it is $24,000 after 1 year, $19,200 after 2 ye

lisabon 2012 [21]

Answer:

It annually drops by 20%

Step-by-step explanation:

30,000 multiplied by 20% is 6000. Subtract 6000 from 30,000 and you get 24,000. 24,000 multiplied by 20% is 4,800. Subtract 4,800 from 24,000 and you get 19,200. Etc.

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3 years ago

Use the distributive property to write an equivalent expression.

wlad13 [49]

Answer:

14x+35y-35

Step-by-step explanation:

distribute the 7 to each term

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2 years ago