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Ask QuestionPosted by Tushar Kaushik 6 years, 10 months ago
- 2 answers
Danush Senthil Kumar 6 years, 10 months ago
Posted by Arun Das 6 years, 10 months ago
- 0 answers
Posted by Atlanta Khanganba 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
The sequence formed by the given numbers is 103,107,111,115, ...,999.
This is an AP in which a = 103 and d = (107 - 103) = 4.
Let the total number of these terms be n. Then,
Tn = 999 {tex}\Rightarrow{/tex} a + (n-1)d = 999
{tex}\Rightarrow{/tex} 103 + (n -1) {tex}\times{/tex} 4 = 999
{tex}\Rightarrow{/tex} (n-1){tex}\times{/tex}4 = 896 {tex}\Rightarrow{/tex} (n-1) = 224 {tex}\Rightarrow{/tex} n = 225.
{tex}\therefore{/tex} middle term = ({tex}\frac{{n + 1}}{2}{/tex})th term = ({tex}\frac{{225 + 1}}{2}{/tex})th term = 113th term.
T113 = (a + 112d) = (103 +112 {tex}\times{/tex} 4) = 551.
{tex}\therefore{/tex} T112 = (551 - 4) = 547.
So, we have to find S112 and (S225 - S113).
Using the formula Sm = {tex}\frac{m}{2}{/tex}(a + l) for each sum, we get
s112 = {tex}\frac{{112}}{2}{/tex}(103+ 547) = (112{tex}\times{/tex}325) = 36400
(S225 - S113) = {tex}\frac{{225}}{2}{/tex}(103 + 999) - {tex}\frac{{113}}{2}{/tex}(103 + 551)
= (225 {tex}\times{/tex} 551)-(113 {tex}\times{/tex} 327)
= 123975 - 36951 =87024.
Sum of all numbers on LHS of the middle term is 36400.
Sum of all numbers on RHS of the middle term is 87024.
Posted by Ansh Sharma 6 years, 10 months ago
- 0 answers
Posted by Ashutosh Savita 6 years, 10 months ago
- 1 answers
Gaurav Seth 6 years, 10 months ago
Σfi = 41+p
Σfixi = 303+9p
Mean = Σfixi / Σfi
7.5 = (303+9p) / (41 +p)
7.5(41 +p) = 303 +9p
307.5 +7.5 p = 303 +9p
307.5 - 303 = 9p - 7.5 p
4.5 = 1.5 p
p = 4.5 / 1.5 = 3
P = 3
Hence, the value of p is 3.
Posted by Ammulesh Kumar 6 years, 10 months ago
- 1 answers
Gaurav Seth 6 years, 10 months ago
Area=1/2×diagonal×sum of offsets
=1/2×30×(6.8+9.6)
=15×16.4
=246cm2
Posted by Aman Kumar 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
Let the radius of original cone be r2
Radius of cut off cone be r1
According to the question
Height of the original cone = 10 cm (given)
The cone is cut off from the midpoint of the height,
therefore, height the cone cut off = 5 cm
{tex}\Delta A O C \sim \Delta AO' D{/tex}
OA = Radius of original cone = r2
O'A' = Radius of cutoff cone = r1
Ratio of radius of two cones = Ratio of the height of cones
{tex}\therefore \quad \frac { A O } { A' O ^ { \prime } } = \frac { r _ { 2 } } { r _ { 1 } } = \frac { 10 } { 5 }{/tex}
{tex}\Rightarrow \quad r _ { 2 } = 2 r _ { 1 }{/tex}
Radius of original cone = 2 (radius of the cut off cone)
Volume of the cut off cone {tex}= \frac { 1 } { 3 } \pi r _ { 1 } ^ { 2 } \times 5{/tex}
{tex}= \frac { 5 } { 3 } \pi r _ { 1 } ^ { 2 }{/tex} cubic units
Volume of original cone {tex}= \frac { 1 } { 3 } \pi \left( 2 r _ { 1 } \right) ^ { 2 } \times 10{/tex}
{tex}= \frac { 40 } { 3 } \pi r _ { 1 } ^ { 2 }{/tex} cubic units
Volume of frustum = Volume of an original cone - Volume of cut off cone
{tex}= \frac { 40 } { 3 } \pi r _ { 1 } ^ { 2 } - \frac { 5 } { 3 } \pi r _ { 1 } ^ { 2 }{/tex}
{tex}= \frac { 35 } { 3 } \pi r _ { 1 } ^ { 2 }{/tex} cubic units
Required ratio = Volume of frustum: Volume of cut off cone
{tex}= \frac { 35 \pi r _ { 1 } ^ { 2 } } { 5 \pi r _ { 1 } ^ { 2 } } = \frac { 7 } { 1 }{/tex}
Therefore, the required ratio = 7: 1.
Posted by Vikram Ranga 6 years, 10 months ago
- 3 answers
Posted by Tanishq Agarwal 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
Given, speed of boat in still water = 8 Km/hr. Let the speed of the stream be x km/hr. Then,
Speed of boat in downstream = (8 + x) km/hr
Speed of boat in upstream = (8 - x) km/hr
We know that time taken to cover 'd' km with speed 's' km/hr is {tex} \frac ds{/tex}
So,Time taken by the boat to go 15 km upstream {tex} = \frac{{15}}{{8 - x}}{/tex} hours.
&, Time taken by the boat to 22 km downstream {tex} = \frac{{22}}{{8 + x}}{/tex} hours.
It is given that the total time taken by boat to go 15 km upstream & 22 km downstream is 5 hours.
{tex}\therefore \frac{{15}}{{8 - x}} + \frac{{22}}{{8 + x}} = 5{/tex}
{tex} \Rightarrow \frac{{15(8 + x) + 22(8 - x)}}{{(8 - x)(8 + x)}} = 5{/tex}
{tex} \Rightarrow \frac{{120 + 15x + 176 - 22x}}{{{8^2} - {x^2}}} = 5{/tex}
{tex} \Rightarrow \frac{{ - 7x + 296}}{{64 - {x^2}}} = 5{/tex}
{tex}\Rightarrow{/tex} -7x + 296 = 5(64 - x2)
{tex}\Rightarrow{/tex} -7x + 296 = 320 - 5x2
{tex}\Rightarrow{/tex} 5x2 - 7x + 296 - 320 = 0
{tex}\Rightarrow{/tex} 5x2 - 7x - 24 = 0
{tex}\Rightarrow{/tex} 5x2 - 15x + 8x - 24 = 0
{tex}\Rightarrow{/tex} 5x(x - 3) + 8(x - 3) = 0
{tex}\Rightarrow{/tex} (5x + 8)(x- 3) = 0
{tex}\Rightarrow{/tex} x - 3 = 0 [{tex}\because{/tex} Speed can not be negative {tex}\therefore{/tex} 5x + 8 {tex}\ne{/tex} 0]
{tex}\Rightarrow{/tex} x = 3
Hence, the speed of the stream is 3 km/hr.
Posted by Rabiya Shadab 6 years, 10 months ago
- 0 answers
Posted by Krishna Sharma 5 years, 8 months ago
- 0 answers
Posted by Quad Squad 6 years, 10 months ago
- 0 answers
Posted by Devkaran Bhagat 6 years, 10 months ago
- 1 answers
Aparna Sinha ?? 6 years, 10 months ago
Posted by Vinod Sharma 6 years, 10 months ago
- 1 answers
Posted by Preeti Rahi 6 years, 10 months ago
- 1 answers
Posted by Amit Shinde 6 years, 10 months ago
- 1 answers
Gaurav Seth 6 years, 10 months ago
By middle term splitting we can write it as,
x2+1200x-800x-960000
x (x+1200)-800(x+1200)
(x-800)(x+1200)
Equating
x-800=0
x=800
x+1200=0
x=-1200
Posted by Sehaj Brar 6 years, 10 months ago
- 1 answers
Gaurav Seth 6 years, 10 months ago
Let :-
a = n³ - n
= n(n² - 1)
= n(n - 1)(n + 1)
= (n - 1)n (n + 1)
1) Now out of three (n - 1),n and (n + 1) one must be even so a is divisible by 2
2) Also (n - 1),n and (n + 1) are three consecutive integers thus as proved a must be divisible by 3
From (1) and (2)
a must be divisible by 2 × 3 = 6
Hence n³ - n is divisible by 6 for any positive integer n
Posted by Last Bencher 6 years, 10 months ago
- 1 answers
Garima Choudhary 6 years, 10 months ago
Posted by Sandeep Gaur 6 years, 10 months ago
- 1 answers
Posted by Ranveer Rathore 6 years, 10 months ago
- 2 answers
Gaurav Seth 6 years, 10 months ago
HCF of 96 and 404 by prime factorisation method:-
Since, 96 = 2 × 2 × 2 × 2 × 2 × 3
and, 404 = 2 × 2 × 101
So, HCF of 96 and 404 = Product of common prime factors
= 2 × 2 = 4
Posted by Santhi Ks 6 years, 10 months ago
- 1 answers
Anjali Kumari? 6 years, 10 months ago
Posted by S K Gowda 6 years, 10 months ago
- 1 answers
Vishakha Singhal 6 years, 10 months ago
Posted by Vishu Rana 6 years, 10 months ago
- 2 answers
Posted by Sahil Karki 6 years, 10 months ago
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Thanks Ki Baarish ........???? 6 years, 10 months ago
Posted by A Kumar 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
Let{tex} AB\ and \ CD{/tex} are towers of height {tex} h_1 \ and \ h_2{/tex} respectively.
If E is the midpoint of BD then {tex}BE = DE = x{/tex}
In right {tex}\triangle{/tex}ABE
{tex}\frac { h _ { 1 } } { x }{/tex}{tex} = tan 60° {/tex}
{tex}\Rightarrow{/tex} h1 = {tex}\sqrt { 3 } x{/tex}
In right {tex}\triangle{/tex}CDE {tex}\frac { h _ { 2 } } { x }{/tex}= tan 30° {tex}\Rightarrow h _ { 2 } = \frac { x } { \sqrt { 3 } }{/tex}
Now {tex}\frac { h _ { 1 } } { h _ { 2 } } = \frac { \sqrt { 3 } x } { \frac { x } { \sqrt { 3 } } } = \frac { 3 } { 1 }{/tex} {tex}\Rightarrow h _ { 1 } : h _ { 2 } = 3 : 1{/tex}
Posted by Hello Of Cbsce 6 years, 10 months ago
- 3 answers
Garima Choudhary 6 years, 10 months ago
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