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Ask QuestionPosted by Debayan Halder 8 years, 8 months ago
- 1 answers
Posted by Zahid Zafir 8 years, 8 months ago
- 2 answers
Rupender Singh 8 years, 8 months ago
To find 5 rational number between 5 & 6, we can write them with denominator 5 + 1 = 6
{tex}5 = {30 \over 6}{/tex}and
{tex}6 = {36 \over 6}{/tex}
Thus, 5 rational no. between 5 & 6 (i.e. {tex}{30 \over 6} and {36 \over 6} {/tex}are
{tex}{31 \over 6}, {32 \over 6}, {33 \over 6}, {34 \over 6}, {35 \over 6}{/tex}
Posted by Anubhav Chamoli 8 years, 8 months ago
- 1 answers
Posted by Parvendra Chouhan 8 years, 8 months ago
- 1 answers
Payal Singh 8 years, 8 months ago
Solution : {tex}{4-3\sqrt 2\over 4+3\sqrt 2}{/tex}
Rationalise it denominator,
{tex}= {4-3\sqrt 2\over 4+3\sqrt 2} \times {4-3\sqrt 2\over 4-3\sqrt 2}{/tex}
{tex}= {(4-3\sqrt 2)^2\over( 4)^2-(3\sqrt 2)^2}{/tex} using (a+b)(a-b) = a2 - b2
{tex}= {16-24\sqrt 2 + 18\over 16-18} {/tex}
{tex}= {34-24\sqrt 2\over -2}{/tex}
{tex}= {-17 +12\sqrt 2}{/tex}
Posted by Eswar Eswar 8 years, 8 months ago
- 1 answers
Naveen Sharma 8 years, 8 months ago
Ans.
Let age of brother = x
Age of sister = y
According to question,
{tex}=> {x\over y} = {3\over 2}{/tex} => 2x = 3y
=> 2x - 3y = 0
Required Linear Equation
Posted by Harshith Kambanna 8 years, 8 months ago
- 1 answers
Naveen Sharma 8 years, 8 months ago
Ans. Let radius of first cylinder = 2x
Height of first cylinder = 5y
Volume of first cylinder = {tex}\pi \times (2x)^2\times 5y = 20\pi x^2y{/tex}
Curved surface area of first cylinder = {tex}2\pi\times 2x\times 5y = 20\pi xy{/tex}
Radius of 2nd cylinder = 3x
Height of 2nd cylinder = 3y
Volume of 2nd cylinder = {tex}\pi (3x)^2 \times 3y = 27\pi x^2y{/tex}
Curved surface Area of 2nd cylinder = {tex}2\pi\times (3x)\times 3y = 18\pi xy{/tex}
Ratio of Volume = {tex}20\pi x^2y\over 27\pi x^2y {/tex}= 20:27
Ratio of CSA = {tex}20\pi xy\over 18\pi xy{/tex}= 10:9
Posted by Gautam Rana 8 years, 8 months ago
- 0 answers
Posted by Hrutvi Thaker 8 years, 8 months ago
- 1 answers
Rashmi Bajpayee 8 years, 8 months ago
Using Heron's formula
s = (35 + 54 + 61)/2 = 75 cm
Area of trianlge = {s(s - a)(s - b)(s - c)}1/2
= {75(75 - 35)(75 - 54)(75 - 61)}1/2
= {75 × 40 × 21 × 14}1/2
= 210√10 sq.cm
Now Area of triangle = 1/2 × base x altitude
210√10 = 1/2 × 35 × altitude
Altitude = 12√10 cm
Posted by Sooraj Gupta 8 years, 8 months ago
- 0 answers
Posted by Tanvi Kausar 8 years, 8 months ago
- 1 answers
Jd Shukla 8 years, 8 months ago
See,for x=0 and y= 0 the x axis and y axis are representators of the equations. and the area enclosed is 4.5 units2
Posted by Jd Shukla 8 years, 8 months ago
- 1 answers
Anjana Krishnan 8 years, 8 months ago
The sum of measures of the interior angles of a polygon is (n - 2)180 and the sum of the exterior angles of a polygon is always 360degree.
Posted by Jd Shukla 8 years, 8 months ago
- 0 answers
Posted by Jd Shukla 8 years, 8 months ago
- 1 answers
Renu Yadav 8 years, 8 months ago
He'll
This is a simple answer what we have to do is
1800 - 1600 = 200
So, 3200 + 200 = 3400
Thanks
Posted by Hardik . 8 years, 8 months ago
- 0 answers
Posted by Vishal Bansal 8 years, 8 months ago
- 1 answers
Posted by Subitha Baskar 8 years, 8 months ago
- 2 answers
Mohit Sharma 7 years, 10 months ago
Given: PQRST is pentagon. TX || SP And RY || SQ (Please refer to the attachment)
RTP: ar(PQRST) = ar(SXY)
Proof:
ΔSPX and ΔSPY are triangles on the same base SP and between same parallels TX and SP because TX || SP
Therefore, ar(ΔSPX) =ar(ΔSPT)
ΔSPX and ΔSPT are triangles on the same base SP and between same parallels TX and SP (Given: TX || SP)
Therefore, ar(ΔSPX) =ar(ΔSPT)
ΔSQY and ΔSQR are triangles on the same base SQ and between same parallels RY and SQ (Given: RY || SQ)
Therefore, ar(ΔSQY) =ar(ΔSQR)
And,
ar(ΔSXY)
=ar(ΔSPX) + ar(ΔSQY) + ar(ΔSPQ)
ar(PQRST)
= ar(ΔSPT) + ar(ΔSQR)+ ar(ΔSPQ)
Plug in ar(ΔSPT)=ar(ΔSPX) and ar(ΔSQR)=ar(ΔSQY):
=ar(ΔSPX) + ar(ΔSQY) + ar(ΔSPQ)
=ar(ΔSXY)
Therefore, ar(PQRST) = ar(SXY) (Proved)
Posted by K. Sai Kishor 8 years, 9 months ago
- 1 answers
Posted by Viji Venkat 8 years, 9 months ago
- 2 answers
Naveen Sharma 8 years, 8 months ago
Ans. Let Hydrogen = x
Oxygen = y
Then According To Ques.
{tex}=> {x\over y} = {2\over 1}{/tex}
=> x = 2y
=> x - 2y = 0
Required Equation.
If y = 6
then x - 2(6)= 0
=> x - 12 = 0
=> x = 12
Hydrogen = 12 g
Posted by Sakuntala Sharma 8 years, 9 months ago
- 1 answers
Naveen Sharma 8 years, 9 months ago
Ans. Let O is center of circle. AB is chord.
So OA is radius. Draw OM perpendicular to AB.
We Know perpendicular from center to the chord bisects it.
So AM = 4cm
Now, In triangle OAM
=> OA2 = AM2 + OM2 .....(Using Pythagoras)
=> 25 = 16 + OM2
=> OM2 = 9
=> OM = 3
Distance of chord from center is 3 cm
Posted by Aarohi Mehra 8 years, 9 months ago
- 1 answers
Naveen Sharma 8 years, 9 months ago
Ans. Deepika gave piece of cloth of area 551m2. Using this cloth she makes a conical tent of radius 7 m and during this process 1m2 of cloth got wasted.
Therefore, tent is made from cloth of area = 551m2 - 1 m2 = 550 m2
The above area being the curved surface area of the tent.
Now, we know that curved surface area of cone ={tex}\pi rl{/tex}
{tex}=> {22\over 7}\times 7\times l= 550{/tex}
=> l = 25m
We know,
{tex}=> l^2 = r^2+h^2{/tex}
{tex}=> 625 = 49 + h^2{/tex}
=> h2 = 576
=> h = 24
Volume of conical tent = {tex}{1\over 3}\pi r^2h{/tex}
{tex}=> {1\over 3} \times {22\over 7}\times 49\times 24 {/tex}
=> 1232m3
Posted by Arpit Kumar 8 years, 9 months ago
- 1 answers
Naveen Sharma 8 years, 9 months ago
Ans. Lateral Surface Area of Cube = 4a2
Volume of cube =a3
Where a is side of cube.
Posted by Guna Kaviya 8 years, 9 months ago
- 1 answers
Naveen Sharma 8 years, 9 months ago
Ans.
Length of rectangular sheet = 16cm
Breadth of rectangular sheet = 3.5cm
When we rotate a rectangle about its one side, the solid thus formed is cylinder.
height of cylinder formed = 16cm
Radius of cylinder formed = {tex}{3.5\over 2\pi}{/tex}
Total Surface Area of cylinder = {tex}2\times \pi \times{ 3.5\over 2\pi}\times 16= 56cm^2{/tex}
Posted by Vishrwa Tiwari 4 years, 11 months ago
- 1 answers
Arnol Jamatia 4 years, 9 months ago
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Rupender Singh 8 years, 8 months ago
To find 5 rational number between 6 and 7, we can write 6 and 7 with denominator 5 + 1 = 6
{tex}6 = {36 \over 6}{/tex}and
{tex}7 = {42 \over 6}{/tex}
Thus, 5 ration number between 6 & 7 (i.e. {tex}{36 \over 6} and {42\over 6}{/tex}) are
{tex}{37 \over 6}, {38\over 6}, {39\over 6}, {40\over 6}, {41 \over 6}{/tex}
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