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andreev551 [17]
2 years ago
7

Find the area of a cylinder of radius 4 cm and height 5 cm

Mathematics
2 answers:
irakobra [83]2 years ago
7 0
<h3>✠ Given:-</h3>

  • Radius of the Cylinder = 4 cm
  • Height of the Cylinder = 5 cm

<h3>✠ To Find:-</h3>

  • Area of the Sphere = ??

<h3>✠ Solution:-</h3>

⚘ Here , we Have Given That Radius of the Sphere is 4 cm and Height of the Cylinder is 5 cm. we have to find the Area of the Cylinder , We know that formula used to find the area; <u>Area of Cylinder = 2πr²+2πrh</u>.....

☼ <u>Calculating the Area of the Cylinder</u>;

\\  \sf \implies \: Area_{ \red{ \{Cylinder \}}} \:  =  \:  \: 2 \: \pi \: r {}^{2} + 2 \: \pi \: r  \: h  \\

\\  \sf \implies \: Area_{ \red{ \{Cylinder \}}} \:  =  \bigg( \: 2 \:  \times  \: \frac{22}{7}  \:  \times \: 4{}^{2} \bigg)+ \bigg(2 \:  \times  \: \frac{22}{7}  \:  \times \: 4 \: \times   \:  5 \bigg)\\

\\  \sf \implies \: Area_{ \red{ \{Cylinder \}}} \:  =  \bigg( \: 2 \:  \times  \: \frac{22}{7}  \:  \times \: 4 \:  \times \:  4 \bigg)+ \bigg(2 \:  \times  \: \frac{22}{7}  \:  \times \: 4 \: \times   \:  5 \bigg)\\

\\  \sf \implies \: Area_{ \red{ \{Cylinder \}}} \:  =  \bigg( \: 2 \:  \times  \: \frac{22}{7}  \:  \times \: 16 \bigg)+ \bigg(2 \:  \times  \: \frac{22}{7}  \:  \times \: 20\bigg)\\

\\  \sf \implies \: Area_{ \red{ \{Cylinder \}}} \:  =  \bigg(  \: \frac{2 \times 22 \times 16}{7}\bigg)+ \bigg(  \: \frac{2 \times 22 \times 20}{7}  \bigg)\\

\\  \sf \implies \: Area_{ \red{ \{Cylinder \}}} \:  =  \bigg(  \: \frac{704}{7}\bigg)+ \bigg(  \: \frac{880}{7}  \bigg)\\

\\  \sf \implies \: Area_{ \red{ \{Cylinder \}}} \:  =  \bigg(  \: \frac{704}{7}\bigg)+ \bigg(  \: \frac{880}{7}  \bigg)\\

\\  \sf \implies \: Area_{ \red{ \{Cylinder \}}} \:  =  100.57 \:  \: +125.7  \:  \: \\

\\  \sf \implies \: Area_{ \red{ \{Cylinder \}}} \:  =  226  \:  cm\: \\

\\  \\ ✠  \:  \: { \underline{\bf { \color{blue}{ \:  \: Additional  \:  \: Information :- \:  \: }}}} \\  \\

⇥ Perimeter of rectangle = 2(length× breadth)

⇥ Diagonal of rectangle = √(length ²+breadth ²)

⇥ Area of square = side²

⇥ Perimeter of square = 4× side

⇥ Volume of cylinder = πr²h

⇥ T.S.A of cylinder = 2πrh + 2πr²

⇥ Volume of cone = ⅓ πr²h

⇥ C.S.A of cone = πrl

⇥ T.S.A of cone = πrl + πr²

⇥ Volume of cuboid = l × b × h

⇥ C.S.A of cuboid = 2(l + b)h

⇥ T.S.A of cuboid = 2(lb + bh + lh)

⇥ C.S.A of cube = 4a²

⇥ T.S.A of cube = 6a²

⇥ Volume of cube = a³

⇥ Volume of sphere = 4/3πr³

⇥ Surface area of sphere = 4πr²

⇥ Volume of hemisphere = ⅔ πr³

⇥ C.S.A of hemisphere = 2πr²

⇥ T.S.A of hemisphere = 3πr²

olga55 [171]2 years ago
4 0

Answer:

226.19cm^2

Step-by-step explanation:

Total surface area of cylinder = 2πr(r + h).

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(a) How many integers from 1 through 1,000 are multiples of 2 or multiples of 9?
DENIUS [597]

Answer:

(a)There are 500 integers from 1 through 1,000 are multiples of 2.

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)0.611

Step-by-step explanation:

Given the set of Integers from 1 through 1000

The least multiple of 2 is 2 and the highest multiple of 2 in the interval is 1000.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 2,4,6,... is an Arithmetic Progression,

where first term, a =2, common difference, d =2

Nth term of an A.P,

U_n= a+(n-1)d\\1000=2+2(n-1)\\1000=2+2n-2\\1000=2n\\n=500

  • There are 500 integers from 1 through 1,000 are multiples of 2.

Similarly,

Given the set of Integers from 1 through 1000

The least multiple of 9 is 9 and the highest multiple of 9 in the interval is 999.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 9,18,27,... is an Arithmetic Progression,

where first term, a =9, common difference, d =9

Nth term of an A.P,

U_n= a+(n-1)d\\999=9+9(n-1)\\999=9+9n-9\\999=9n\\n=111

  • There are 111 integers from 1 through 1,000 are multiples of 9.

(b)Probability that an integer selected at random is a multiple of 2 or a multiple of 9.

There are a total of 1000 numbers between from 1 through 1,000

n(S)=1000

n(Multiples of 2)=500

n(Multiples of 9)=111

Therefore:

P(\text{Multiples of 9}) \:OR\: P(\text{Multiples of 2})=\frac{111}{1000} + \frac{500}{1000} \\=\frac{611}{1000} \\=0.611

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If A = {2, 3, 4, 5} and B = {5, 6, 7, 8}, what is A U B?<br>{2, 3, 4, 5, 6, 7, 8)<br>{5}​
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<em>Hey</em><em>!</em><em>!</em><em>!</em><em>!</em>

<em>Here</em><em>'s</em><em> </em><em>you</em><em>r</em><em> </em><em>answer</em><em>:</em>

<em>Sol</em><em>ution</em><em>:</em>

<em>A</em><em>=</em><em>{</em><em>2</em><em>,</em><em>3</em><em>,</em><em>4</em><em>,</em><em>5</em><em>}</em>

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<em> </em><em>A</em><em> </em><em>U</em><em> </em><em>B</em><em>=</em><em>{</em><em>2</em><em>,</em><em>3</em><em>,</em><em>4</em><em>,</em><em>5</em><em>}</em><em> </em><em>U</em><em> </em><em>{</em><em>5</em><em>,</em><em>6</em><em>,</em><em>7</em><em>,</em><em>8</em><em>}</em>

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<em>In</em><em> </em><em>case</em><em> </em><em>of</em><em> </em><em>Union</em><em>,</em><em>we</em><em> </em><em>have</em><em> </em><em>to</em><em> </em><em>list</em><em> </em><em>all</em><em> </em><em>the</em><em> </em><em>elements</em><em> </em><em>which</em><em> </em><em>are</em><em> </em><em>present</em><em> </em><em>in</em><em> </em><em>both</em><em> </em><em>sets</em><em>.</em>

<em>Hope</em><em> </em><em>it</em><em> </em><em>helps.</em><em>.</em><em>.</em>

<em>Good</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>

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