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Ask QuestionPosted by Sagar Rastogi 6 years, 10 months ago
- 1 answers
Posted by Gungun Mahajan 6 years, 10 months ago
- 0 answers
Posted by Uvi Prajapati 6 years, 10 months ago
- 1 answers
Yogita Ingle 6 years, 10 months ago
We know that a x b = H.c.f (a, b) x L.c.m( a,b)
a = 45 , H.c.f (a, b) = 9 ,L.c.m( a,b) = 360.
b = ( 9 x 360) / 45 = 72.
∴ Another number is 72
Posted by Aditya Singh 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
In the figure, D is a point on side BC of {tex}\triangle {/tex} ABC such that.{tex}\frac{{BD}}{{CD}} = \frac{{AB}}{{AC}}{/tex}
To prove, AD is the bisector of {tex}\angle{/tex} BAC
Construction: From BA produce cut off AE = A. Join CE
Proof:{tex}\frac{{BD}}{{CD}} = \frac{{AB}}{{AC}}{/tex} ........Given
{tex}\Rightarrow {/tex} {tex}\frac{{BD}}{{CD}} = \frac{{AB}}{{AE}}{/tex} {tex}\because {/tex} AC = AE(by construction)
{tex}\therefore {/tex} In {tex}\triangle {/tex} BCE,
AD {tex}\parallel{/tex} CE ............By converse of the basic proportional theorem
{tex}\therefore {/tex} {tex}\angle{/tex}BAD = {tex}\angle{/tex}AEC .......(1)...........Corres. {tex}\angle{/tex} s
{tex}\angle{/tex}CAD = {tex}\angle{/tex}AEC ..........(2) ........Alt,Int. {tex}\angle{/tex} s
{tex}\therefore {/tex} AC = AE .........By construction
{tex}\therefore {/tex} {tex}\angle{/tex}AEC = {tex}\angle{/tex}ACE ............(3)...angles opposite equal sides of a triangle are equal
Using (3),(1) and (2) gives {tex}\angle{/tex}BAD = {tex}\angle{/tex}CAD
Posted by Yash Sharma ?? 6 years, 10 months ago
- 3 answers
Posted by Ramakant??? 8306360538 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
According to the question,
{tex}\frac { 2 } { 3 } \operatorname { cosec } ^ { 2 } 58 ^ { \circ } - \frac { 2 } { 3 } \cot 58 ^ { \circ } \tan 32 ^ { \circ } - \frac { 5 } { 3 }{/tex}{tex} tan13° tan37° tan45° tan53° tan77°{/tex}
= {tex}\frac { 2 } { 3 } \operatorname { cosec } ^ { 2 } 58 ^ { \circ } - \frac { 2 } { 3 } \cot 58 ^ { \circ } \tan \left( 90 ^ { \circ } - 58 ^ { \circ } \right) - \frac { 5 } { 3 }{/tex}tan 13° tan 37° tan 45°tan (90° - 37°) tan (90° -13°)
= {tex}\frac { 2 } { 3 } \operatorname { cosec } ^ { 2 } 58^\circ - \frac { 2 } { 3 } \cot ^ { 2 } 58^\circ - \frac { 5 } { 3 }{/tex}{tex}tan 13° tan37° tan 45° cot 37° cot 13°{/tex}
= {tex}\frac { 2 } { 3 } \left( \operatorname { cosec } ^ { 2 } 58 ^ { \circ } - \cot ^ { 2 } 58 ^ { \circ } \right) - \frac { 5 } { 3 } \tan 13 ^ { \circ } \tan 37 ^ { \circ } \times 1 \times \frac { 1 } { \tan 37 ^ { \circ } } \times \frac { 1 } { \tan 13 ^ { \circ } }{/tex}
= {tex}\frac { 2 } { 3 } \times 1 - \frac { 5 } { 3 } = \frac { 2 } { 3 } - \frac { 5 } { 3 } = -1{/tex}
Posted by Mukul Empire 6 years, 10 months ago
- 2 answers
Gaurav Seth 6 years, 10 months ago
Let numbers be x at ones place and y at tens place so
10y+x is that digit
now reversed digit is 10x+y
according to question
7 (10y+x)=4(10x+y)
x=2y (i)
now
given
x-y=3
from eq. i 2y=x
2y-y=3
y=3
so
x=63
required original no. is 36 and reversed digit is 63
Posted by Shardul Ganorkar 6 years, 10 months ago
- 2 answers
Gaurav Seth 6 years, 10 months ago
Let father = x, son = y
x = 3y + 3
x + 3 = 2(y + 3) + 10
3y + 3 + 3 = 2y + 6 + 10
3y + 6 = 2y + 16
3y - 2y = 16 - 6
y = 10
x = 3y + 3
x = 3 × 10 + 3
x = 30 + 3
x = 33
The present age of the father is 33 years old
Posted by Ananya Sharma 6 years, 10 months ago
- 2 answers
Posted by Nitish Sharma 6 years, 10 months ago
- 1 answers
Mukul Empire 6 years, 10 months ago
Posted by Adit Arab 6 years, 10 months ago
- 1 answers
Posted by Sara Sharma 6 years, 10 months ago
- 3 answers
Maniti Gupta 6 years, 10 months ago
Dia Khurana@1608 6 years, 10 months ago
Posted by Spandana D 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
According to question, a=2 and {tex}\mathrm { S } _ { 5 } = \frac { 1 } { 4 } \left[ \mathrm { S } _ { 10 } - \mathrm { S } _ { 5 } \right]{/tex}
{tex}\Rightarrow{/tex} 4S5 = S10 - S5
{tex}\Rightarrow{/tex} 5S5 = S10
{tex}\Rightarrow \sqrt [ 5 ] { \frac { 5 } { 2 } \{ 2 \times 2 + ( 5 - 1 ) d \} } ] = \frac { 10 } { 2 } [ 2 \times 2 + ( 10 - 1 ) d ]{/tex} since {tex}{S_n} = {n \over 2}\left[ {2a + (n - 1)d} \right]{/tex}
{tex}\Rightarrow \frac { 25 } { 2 } [ 4 + 4 d ] = 5 [ 4 + 9 d ]{/tex}
{tex}\Rightarrow{/tex} 25[4 + 4d] = 10[4 + 9d]
{tex}\Rightarrow{/tex} 100 + 100d = 40 + 90d
{tex}\Rightarrow{/tex} 10d = -60
{tex}\Rightarrow{/tex} d = -6
Now, an = a + (n - 1)d
{tex}\Rightarrow a _ { 20 } = 2 + ( 20 - 1 ) \times ( - 6 ){/tex}
{tex}\Rightarrow{/tex} a20 = 2 - 114 = -112
Posted by Sanjana Gupta 6 years, 10 months ago
- 1 answers
Posted by Vaishnavi Barde 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Here, prime factorization of N = 2³×54 ×81×7
=23×54× 81 x 7
=23×54×81 x 7
= (10)3×5 x 81 x 7
So, number of consecutive zeroes in 23×53 that is in (10)3 is equal to 3
Posted by Kapil Pandit 6 years, 10 months ago
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Posted by Affu 😊 6 years, 10 months ago
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Posted by Kunalgarg Kunakgarg 6 years, 10 months ago
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Posted by Shruti Thalor 6 years, 10 months ago
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Posted by Suryansh Singh 6 years, 10 months ago
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Farhat Shaikh Shaikh 6 years, 10 months ago
Abhinav Kumar 6 years, 10 months ago
Rakesh Vyas 6 years, 10 months ago
Posted by Suryansh Singh 6 years, 10 months ago
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Anureet Kaur 6 years, 10 months ago
Posted by Prince Garg 6 years, 10 months ago
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Posted by Shubhagani Verma 6 years, 10 months ago
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Khushboo Giri 6 years, 10 months ago
Posted by Ajay Gaur 6 years, 10 months ago
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Posted by Xyz Abc 6 years, 10 months ago
- 3 answers
Ram Kushwah 6 years, 10 months ago
{tex}\begin{array}{l}\text{240 m/s=}\frac{\displaystyle240/1000\;(\mathrm m\;\mathrm{to}\;\mathrm{km})}{1/3600\;(\mathrm s\;\mathrm{to}\;\mathrm{hr})}\\=\frac{\displaystyle240}{1000}\times\frac{3600}1\\=24\times36=860\;\mathrm{km}/\mathrm{hr}\end{array}{/tex}
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Geetanand Yadav 6 years, 10 months ago
4Thank You