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Ask QuestionPosted by Punitha Punitha 6 years, 11 months ago
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Posted by Rahul Raj 6 years, 11 months ago
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Puja Sahoo? 6 years, 11 months ago
Posted by Saurabh Singh Lodhi 6 years, 11 months ago
- 7 answers
Yogita Ingle 6 years, 11 months ago
Let one of the numbers be x.
Other number is (15 - x)
Sum of their reciprocals = 3/10
1/x + 1/(15 - x) = 3/10
(15 ) / (15x - x2) = 3/10
150 = 45x - 3x2
3x2 - 45x + 150 = 0
x2 - 15x + 50 = 0
(x - 10)(x - 5) = 0
x = 10 or 5
Therefore, the numbers are 10 and 5.
If x = 15
Puja Sahoo? 6 years, 11 months ago
Posted by Amrit Panag Panag 6 years, 11 months ago
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Rahul Raj 6 years, 11 months ago
Susmita Mandal 6 years, 11 months ago
Posted by Jai Kishan 6 years, 11 months ago
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Rahul Raj 6 years, 11 months ago
Ragini Jha? 6 years, 11 months ago
Posted by Ricky Roy 6 years, 11 months ago
- 4 answers
Gaurav Seth 6 years, 11 months ago
Let BC be the building and AD be the tower.
Let the height of tower, AD be h m.
Angles of depression of the top D and the bottom A of the tower CB are 30° and 60° respectively.
∴ ∠CDE = 30°
∠CAB = 60°
Since, BC = 60 m.
∴ CE = (60 – h) m
Let AB = DE = x m
In ∆DEC,
In ∆CBA,
Equating equation (1) and (2),
⇒ 3 (60 – h) = 60
⇒ 180 – 3h) = 60
⇒ 180 – 60 = 3n
⇒ 120 = 3h
⇒ h = 40
Thus, the height of the tower is 40 m.
Posted by Chetan Pal 6 years, 11 months ago
- 3 answers
Yogita Ingle 6 years, 11 months ago
a3 = a + (3 – 1)d = a + 2d = 5 ... (1)
a7 = a + (7 – 1)d = a + 6d = 9 ... (2)
Solving the pair of linear equations (1) and (2),
a = 3, d = 1
Hence, the required AP is 3, 4, 5, 6, 7, ...
Posted by Darshan Dhanu 6 years, 11 months ago
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Posted by Rajni Shah 6 years, 11 months ago
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Posted by Shyam Agarawal 6 years, 11 months ago
- 5 answers
Abhi Rwt 6 years, 11 months ago
Rahul Raj 6 years, 11 months ago
Rahul Raj 6 years, 11 months ago
Posted by Tarun Agrawal 6 years, 11 months ago
- 5 answers
Priyanshu Singh 6 years, 11 months ago
Posted by Ilakya S 6 years, 11 months ago
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Prince Rajput 6 years, 11 months ago
Posted by Harpreet Chahal 6 years, 11 months ago
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Rahul Raj 6 years, 11 months ago
Raghubir Singh 6 years, 11 months ago
Posted by Harmanjot Singh 6 years, 11 months ago
- 2 answers
Rahul Raj 6 years, 11 months ago
Yash Patil 6 years, 11 months ago
Posted by Shivam Singh 6 years, 11 months ago
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Posted by Jogendrabhoi Jogendrabhoi 6 years, 11 months ago
- 3 answers
Priyanshu Singh 6 years, 11 months ago
Vaibhav Baliyan 6 years, 11 months ago
Light Yagami 6 years, 11 months ago
Posted by Shivam Singh 6 years, 11 months ago
- 1 answers
Gaurav Seth 6 years, 11 months ago
Let present age of Aftab be x
And, present age of daughter is represented byy
Then Seven years ago,
Age of Aftab = x -7
Age of daughter = y-7
According to the question,
(x - 7) = 7 (y – 7 )
x – 7 = 7 y – 49
x- 7y = - 49 + 7
x – 7y = - 42 …(i)
x = 7y – 42
Putting y = 5, 6 and 7, we get
x = 7 × 5 - 42 = 35 - 42 = - 7
x = 7 × 6 - 42 = 42 – 42 = 0
x = 7 × 7 – 42 = 49 – 42 = 7
x-707y567
Three years from now ,
Age of Aftab = x +3
Age of daughter = y +3
According to the question,
(x + 3) = 3 (y + 3)
x + 3 = 3y + 9
x -3y = 9-3
x -3y = 6 …(ii)
x = 3y + 6
Putting, y = -2,-1 and 0, we get
x = 3 × - 2 + 6 = -6 + 6 =0
x = 3 × - 1 + 6 = -3 + 6 = 3
x = 3 × 0 + 6 = 0 + 6 = 6
| <div align="center" style="margin:0in 0in 0in 40.5pt; text-align:center; padding:0in 5.4pt">X</div> | <div align="center" style="margin:0in 0in 0in 40.5pt; text-align:center; padding:0in 5.4pt">0</div> | <div align="center" style="margin:0in 0in 0in 40.5pt; text-align:center; padding:0in 5.4pt">3</div> | <div align="center" style="margin:0in 0in 0in 40.5pt; text-align:center; padding:0in 5.4pt">6</div> |
| <div align="center" style="margin:0in 0in 0in 40.5pt; text-align:center; padding:0in 5.4pt">y</div> | <div align="center" style="margin:0in 0in 0in 40.5pt; text-align:center; padding:0in 5.4pt">-2</div> | <div align="center" style="margin:0in 0in 0in 40.5pt; text-align:center; padding:0in 5.4pt">-1</div> | <div align="center" style="margin:0in 0in 0in 40.5pt; text-align:center; padding:0in 5.4pt">0</div> |
From equation (i) and (ii)
x – 7y = – 42 …(i)
x - 3y = 6 …(ii)</div>
Posted by Khushi Kus 6 years, 11 months ago
- 2 answers
Gaurav Seth 6 years, 11 months ago
Q.Cot A+cosecA=5 Find cosA
<hr />Here is the required solution:
<hr />
Posted by Khushi Kus 6 years, 11 months ago
- 0 answers
Posted by Upneet Chandi 6 years, 11 months ago
- 1 answers
Gaurav Seth 6 years, 11 months ago
Q.Find the value of k for which (a-3b) is a factor of a4-7a2b2+kb4. Hence for this value of k, factorize a4-7a2b2+kb4 .
<hr />Step-by-step explanation:
Posted by Jatin Chawla 6 years, 11 months ago
- 2 answers
Posted by Rohit Rohit 6 years, 11 months ago
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Posted by Rohit Rohit 6 years, 11 months ago
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Posted by Rahul Pal 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Length of the arc =
{tex}\frac { \theta \pi \times r } { 180 ^ { \circ } } = \frac { 120 } { 180 } \times \frac { 22 } { 7 } \times 12{/tex}= circumference of the base of the cone
Let radius of the cone be r.
{tex}\Rightarrow \quad 2 \times \pi \times r = \frac { 120 } { 180 } \times \frac { 22 } { 7 } \times 12{/tex}{tex}\Rightarrow \quad r = \frac { 2 } { 3 } \times \frac { 12 } { 2 } = 4 \mathrm { cm }{/tex}
r = 4 cm, l = 12 cm
{tex}\Rightarrow{/tex} h2 = l2 - r2 = 122 - 42 = 114 - 16
h2 = 128 {tex}\Rightarrow h = \sqrt { 128 } = 8 \sqrt { 2 }{/tex}cm
Volume of the cone = {tex}\frac { 1 } { 3 } \times \pi \times r ^ { 2 } \times h{/tex}
{tex}= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times ( 4 ) ^ { 2 } \times 8 \times \sqrt { 2 }{/tex}
{tex}= \frac { 1 } { 3 } \times \frac { 22 } { 7 } \times 16 \times 8 \times 1.414 \mathrm { cm } ^ { 3 }{/tex}
= 189.61 cm3
Posted by Nitin Mishra 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Let A.P be a , a + d, a + 2d, ...
a is the first term, d is the common difference
The formula of sum of n terms is
{tex}S_{n}=\frac{n}{2}[2 a+(n-1) d]{/tex}
Where n is the number of terms
The sum of its first 10 term is - 80. Therefore,
{tex}-80=\frac{10}{2}[2 a+(10-1) d]{/tex}
-80 = 5[2a + 9d]
-16 = 2a + 9d ...(1)
Sum of its next 10 terms is -280.
{tex}S_{20}=\frac{20}{2}[2 a+(20-1) d]{/tex}
{tex}S_{20}{/tex} = 10[2a + 19d]
Since sum of the next 10 terms of AP is -280. Therefore,
{tex}S_{20}-S_{10}{/tex} =-280
10[2a + 19d] - (-80) = -280
10[2a + 19d] = 360
2a + 19d = -36 ...(2)
Subtract (1) from (2)
2a + 19d - 2a - 9d = -36 + 16
10d = -20
d = -2
Put value of d in (1)
-16 = 2a + 9d
-16 = 2a + 9(-2)
-16 = 2a - 18
2 = 2a
a = 1
Therefore, The AP series is 1, -1, -3, -5 ...
Posted by Nitin Mishra 6 years, 11 months ago
- 0 answers
Posted by Jatin Chawla 6 years, 11 months ago
- 1 answers
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Puja Sahoo? 6 years, 11 months ago
1Thank You