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Ask QuestionPosted by Omkar Dubey 6 years, 10 months ago
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Posted by Vishal Vishu K. Mathur 6 years, 10 months ago
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Posted by Vishal Vishu K. Mathur 6 years, 7 months ago
- 1 answers
Sia ? 6 years, 7 months ago
Here we are having a=3
and d=15-3=12
Let nth term Tn =T21+120
so a+(n-1)d = a+20d+120
or (n-1){tex}\times{/tex}12 = 20 {tex}\times{/tex} 12+120
12n -12 = 240+120
12n = 372
{tex}\style{font-size:12px}{\text{n=}\frac{372}{12}=31}{/tex}
Hence, 31st term is the required term.
Posted by Vishal Vishu K. Mathur 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
Let P(a, 0) be the point which divides the line segment joining A(1, -3) and B(4,5) in ratio K: 1
Using section formula we get,
{tex}(a,0)=( \frac{K\times4+1\times 1}{K+1}, \frac{K\times5+1\times-3}{K+1}){/tex}
{tex}(a,0)=( \frac{4K+1}{K+1}, \frac{5K-3}{K+1}){/tex}
a = {tex}\frac{4 \mathrm{K}+1}{\mathrm{K}+1}{/tex} and 0 = {tex}\frac{5 \mathrm{K}-3}{\mathrm{K}+1}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{5K-3}{K+1} =0{/tex} [ taking y-coordinate]
{tex}\Rightarrow{/tex} 5K-3=0
{tex}\Rightarrow{/tex} K = {tex}\frac{3}{5}{/tex}
The required ratio is 3: 5.
Put the value of k in x-coordinate, we get,
{tex}a =\frac{4 \mathrm{(\frac{3}{5})}+1}{\mathrm{(\frac{3}{5})}+1}{/tex}
{tex}a =\frac{\mathrm{\frac{12+5}{5}}}{\mathrm{\frac{3+5}{5}}} =\frac{\frac{17}{5}}{\frac{8}{5}}=\frac{17}{5}\times \frac{5}{8}=\frac{17}{8}{/tex}
Hence, point P is ({tex}\frac{17}{8}{/tex}, 0).
Posted by Vishal Vishu K. Mathur 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
Compare x2 + 4x + k = 0 with ax2 + bx + c = 0
a = 1 , b = 4, c = k
Since roots of the equation x2 + 4x + k = 0 are real
{tex}\Rightarrow{/tex} b2 - 4ac = (4)2 - 4 (1) (k) = 16 - 4k {tex}\geq{/tex} 0
{tex}\Rightarrow{/tex} k {tex}\leq{/tex} 4
Posted by Muhammad Usman 6 years, 10 months ago
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Posted by Vishal Vishu K. Mathur 6 years, 7 months ago
- 1 answers
Sia ? 6 years, 7 months ago
tan 2A = cot (90° - 2A)
{tex}\therefore{/tex} 90° - 2A = A - 24°
{tex}\Rightarrow{/tex} A = 38°
Posted by S Tiwari 6 years, 10 months ago
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Sia ? 6 years, 4 months ago
Radius of the cone, r2 = 6cm
Height of the cone, h = 24cm
Let the radius of the sphere be 'r1'
Volume of the sphere = Volume of the cone
{tex}\therefore \quad \frac { 4 } { 3 } \pi r _ { 1 } ^ { 3 } = \frac { 1 } { 3 } \pi r _ { 2 } ^ { 2 } h{/tex}
{tex}\frac { 4 } { 3 } \times r _ { 1 } ^ { 3 } = ( 6 ) ^ { 2 } \times \frac { 24 } { 3 }{/tex}
{tex}4 r _ { 1 } ^ { 3 } = 36 \times 24{/tex}
{tex}r _ { 1 } ^ { 3 } = 6 ^ { 3 }{/tex}
{tex}r _ { 1 } = 6{/tex}
Hence, radius of the sphere = 6 cm.
Posted by Ansh Setia 6 years, 7 months ago
- 1 answers
Sia ? 6 years, 7 months ago
According to the question,
Let O be the position of the bird, B be the position of the boy and FG be the building at which G is the position of the girl.
{tex}OL\perp BF \ and\ GM \perp OL.{/tex}
BO = 100m,
{tex}\angle O B L = 30 ^ { \circ },{/tex}
FG = 20m and
{tex}\angle O G M = 45 ^ { \circ }.{/tex}
From right {tex}\Delta O L B,{/tex} we have
{tex}\frac { O L } { B O } = \sin 30 ^ { \circ } \Rightarrow \frac { O L } { 100 \mathrm { m } } = \frac { 1 } { 2 }{/tex}
{tex}\Rightarrow O L = 100 \mathrm { m } \times \frac { 1 } { 2 } = 50 \mathrm { m }{/tex}
{tex}OM = OL - ML \\= OL - FG \\= 50m - 20m \\= 30m.{/tex}
From right {tex}\Delta O M G,{/tex} we have
{tex}\frac { O M } { O G } = \sin 45 ^ { \circ } = \frac { 1 } { \sqrt { 2 } }{/tex}
{tex}\Rightarrow O G = \sqrt { 2 } \times O M \\= \sqrt { 2 } \times 30 \mathrm { m }{/tex}
{tex}=30 \times 1.41 \mathrm { m } \\= 42.3 \mathrm { m }.{/tex}
Distance of the bird from the girl = 42.3m.
Posted by Ankita Rai 6 years, 10 months ago
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Posted by Devendra Chaudhary 6 years, 10 months ago
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Lustig Yashu? 6 years, 10 months ago
Posted by Lustig Yashu? 6 years, 10 months ago
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Posted by Abhinav Raj 6 years, 10 months ago
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Posted by Amogh Anand 6 years, 10 months ago
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Posted by Anuj Patel 6 years, 10 months ago
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Harshita Srivastava 6 years, 10 months ago
Posted by Focult Hixx 6 years, 10 months ago
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Kartik Sharma 6 years, 10 months ago
Posted by Aksh Chhina 6 years, 10 months ago
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Posted by Rajini Prashwanath 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let O be the centre of the given circle.
Then, OP = 10 cm. Also, PT = 8 cm.
Join OT.
Now, PT is a tangent at T and OT is the radius through the point of contact T.
{tex}\therefore \quad O T \perp P T {/tex}
In the right {tex}\Delta O T P {/tex}
we have
OP2 = OT2 + PT2 [by Pythagoras' theorem]
{tex}\Rightarrow{/tex} OT = {tex}\sqrt { O P ^ { 2 } - P T ^ { 2 } } = \sqrt { ( 10 ) ^ { 2 } - ( 8 ) ^ { 2 } } \mathrm { cm } = \sqrt { 36 } \mathrm { cm } {/tex} = 6cm.
Hence, the radius of the circle is 6 cm.
Posted by Rajini Prashwanath 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
The given pair of linear equation is
3x + y = 1, (2k - 1) x + (k - 1) y = 2k + 1
{tex}\Rightarrow{/tex} 3x + y - 1 = 0
Here, (2k - 1)x + (k - 1) y - (2k + 1) = 0
a1 = 3, b1 = 1, c1 = -1
a2 = 2k - 1, b2 = k - 1, c2 = -(2k + 1)
for having no solution, we must have {tex}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}{/tex}
{tex}\Rightarrow \frac{3}{{2k - 1}} = \frac{1}{{k - 1}} \ne \frac{{ - 1}}{{ - (2k + 1)}}{/tex}
From above we have {tex}\frac{3}{{2k - 1}} = \frac{1}{{k - 1}}{/tex}
{tex}\Rightarrow{/tex} 3(k - 1) = 2k - 1
{tex}\Rightarrow{/tex} 3k - 3 = 2k - 1
{tex}\Rightarrow{/tex} 3k - 2k = 3 - 1
{tex}\Rightarrow{/tex} k = 2
Hence, the required value of k is 2.
Posted by Vivek Gupta 6 years, 10 months ago
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Thanks Ki Baarish ........???? 6 years, 10 months ago
Posted by Jatin Dhurwe 6 years, 10 months ago
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Posted by Gurjot Kunder 6 years, 10 months ago
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Thanks Ki Baarish ........???? 6 years, 10 months ago
Posted by Lustig Yashu? 6 years, 10 months ago
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Thanks Ki Baarish ........???? 6 years, 10 months ago
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