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riadik2000 [5.3K]
3 years ago
12

A 3cm x 2cm rectangle sits inside a circle with a radius of 4cm

Mathematics
2 answers:
9966 [12]3 years ago
8 0

The area of the shaded region is 44.24\ cm^{2}44.24 cm2

Step-by-step explanation:

we know that

The area of the shaded region is equal to the area of the circle minus the area of rectangle

step 1

Find the area of circle

The area of the circle is equal to

A=\pi r^{2}A=πr2

we have

r=4\ cmr=4 cm

substitute

A=\pi (4)^{2}A=π(4)2

A=16\pi\ cm^{2}A=16π cm2

step 2

Find the area of rectangle

The area of rectangle is equal to

A=(3)(2)=6\ cm^{2}A=(3)(2)=6 cm2

step 3

Find the difference

16\pi\ cm^{2}-6\ cm^{2}16π cm2−6 cm2

assume

\pi=3.14π=3.14

16(3.14)\ cm^{2}-6\ cm^{2}=44.24\ cm^{2}16(3.14) cm2−6 cm2=44.24 cm2

Alina [70]3 years ago
6 0

Answer:

44,26cm²

Step-by-step explanation:

First find the area of the whole circle using the formula

a = \pi \times  {r}^{2}

a = 3,14 × 4²

= 3,14 × 16

= 50,24cm²

Find the area of the rectangle

r = bh

r = 3 × 2

= 6

Then subtract the area of circle and rectangle

50,24cm² - 6cm² =44,26cm²

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TiliK225 [7]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
What is 0.66666666666 rounded
Oksana_A [137]

Answer:

0.66666666666 as a decimal

How much is the decimal number 0.66666666666 written as (converted to) a percentage? Answer: 66.666666666%

4 0
3 years ago
Read 2 more answers
What is the area of the sector? Either enter an exact answer in terms of \piπpi or use 3.143.143, point, 14 for \piπpi and enter
n200080 [17]

Question:

What is the area of the sector? Either enter an exact answer in terms of π or use 3.14 and enter your answer as a decimal rounded to the nearest hundredth.

Answer:

See Explanation

Step-by-step explanation:

The question is incomplete as the values of radius and central angle are not given.

However, I'll answer the question using the attached figure.

From the attached figure, the radius is 3 unit and the central angle is 120 degrees

The area of a sector is calculated as thus;

Area = \frac{\alpha }{360} * \pi r^2

Where \alpha represents the central angle and r represents the radius

By substituting \alpha = 120 and r = 3

Area = \frac{\alpha }{360} * \pi r^2 becomes

Area = \frac{120}{360} * \pi * 3^2

Area = \frac{1}{3} * \pi * 9

Area = \pi * 3

Area = 3\pi square units

Solving further to leave answer as a decimal; we have to substitute 3.14 for \pi

So, Area = 3\pi becomes

Area = 3 * 3.14

Area = 9.42 square units

Hence, the area of the sector in the attached figure is 3\pi or 9.42 square units

8 0
3 years ago
I need help on math on how to solve a linear equation. So how do you solve (x-2)+(2x+5).
saveliy_v [14]
So, put your "x's" together first and then your intercepts. That should answer your question :)
4 0
2 years ago
given examples of relations that have the following properties 1) relexive in some set A and symmetric but not transitive 2) equ
rodikova [14]

Answer: 1) R = {(a, a), (а,b), (b, a), (b, b), (с, с), (b, с), (с, b)}.

It is clearly not transitive since (a, b) ∈ R and (b, c) ∈ R whilst (a, c) ¢ R. On the other hand, it is reflexive since (x, x) ∈ R for all cases of x: x = a, x = b, and x = c. Likewise, it is symmetric since (а, b) ∈ R and (b, а) ∈ R and (b, с) ∈ R and (c, b) ∈ R.

2) Let S=Z and define R = {(x,y) |x and y have the same parity}

i.e., x and y are either both even or both odd.

The parity relation is an equivalence relation.

a. For any x ∈ Z, x has the same parity as itself, so (x,x) ∈ R.

b. If (x,y) ∈ R, x and y have the same parity, so (y,x) ∈ R.

c. If (x.y) ∈ R, and (y,z) ∈ R, then x and z have the same parity as y, so they have the same parity as each other (if y is odd, both x and z are odd; if y is even, both x and z are even), thus (x,z)∈ R.

3) A reflexive relation is a serial relation but the converse is not true. So, for number 3, a relation that is reflexive but not transitive would also be serial but not transitive, so the relation provided in (1) satisfies this condition.

Step-by-step explanation:

1) By definition,

a) R, a relation in a set X, is reflexive if and only if ∀x∈X, xRx ---> xRx.

That is, x works at the same place of x.

b) R is symmetric if and only if ∀x,y ∈ X, xRy ---> yRx

That is if x works at the same place y, then y works at the same place for x.

c) R is transitive if and only if ∀x,y,z ∈ X, xRy∧yRz ---> xRz

That is, if x works at the same place for y and y works at the same place for z, then x works at the same place for z.

2) An equivalence relation on a set S, is a relation on S which is reflexive, symmetric and transitive.

3) A reflexive relation is a serial relation but the converse is not true. So, for number 3, a relation that is reflexive but not transitive would also be serial and not transitive.

QED!

6 0
3 years ago
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