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Ask QuestionPosted by Suman Suman 4 years, 8 months ago
- 1 answers
Posted by Mitu Khan Sanmukh 4 years, 8 months ago
- 1 answers
Gaurav Seth 4 years, 8 months ago
Given quadratic equation is, px² -2√5 px + 15 = 0
Compare px² -2√5 px + 15 = 0 with ax² + bx + c = 0
a = p , b = -2√5p , c = 15
We know that , If the roots of the quadratic equation are equal , then it's discriminant (D) equals to zero.
discriminant = 0
b² - 4ac = 0
( -2√5 p )² - 4× p × 15 = 0
20p² - 60p = 0
20p ( p - 3 ) = 0
p - 3 = 0
p = 3
Posted by Abhishek Kumar 4 years, 8 months ago
- 2 answers
Gaurav Seth 4 years, 8 months ago
The series is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15
The multiples of 4 in the above series is 4, 8, 12 = 3 terms
The probability of selecting a multiple of 4 =
or = 0.2
Yogita Ingle 4 years, 8 months ago
From Number1,2,3,4,5,6,7,8,9,10,11,12,13,14,15
From 1 to 15, Multiples of 4 are 4, 8, 12 only
So Probability= Count of No. which are multiple of 4/Total Number given
Prob.=3/15 = 1/5
Posted by Tabrez Alam 4 years, 8 months ago
- 1 answers
Gaurav Seth 4 years, 8 months ago
माप, साइज या चौड़ाई
किसी वर्ग अंतराल का ऊपर का मान उस वर्ग अंतराल की उपरि वर्ग सीमा कहलाता है तथा उसका नीचे का मान उसकी निम्न वर्ग सीमा कहलाता है। किसी वर्ग की उपरि वर्ग सीमा और निम्न वर्ग सीमा का अंतर उस वर्ग की माप, साइज या चौड़ाई कहलाता है।
Posted by Tabrez Alam 4 years, 8 months ago
- 1 answers
Posted by Adarsh Tiwari 4 years, 8 months ago
- 1 answers
Vaibhav Shukla 1 year, 8 months ago
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Posted by Lalu Gaming 4 years, 8 months ago
- 1 answers
Gaurav Seth 4 years, 8 months ago
Given :
To Find : solve using substitution method
Solution :
-----1
------2
Substitute the value of x from 1 in 2
So, x = 2 and y = 3
Posted by Lalu Gaming 4 years, 8 months ago
- 0 answers
Posted by Mohit Sharma 4 years, 8 months ago
- 1 answers
Gaurav Seth 4 years, 8 months ago
We will check the result
13 × 1 7 × 19 × 23 + 13 × 23
221 × 437 + 299
96577 + 299
96876
Let's do factorization
96876 = 2 × 2× 3 × 3 × 3× 3 ×13 × 23
It is divisible by numbers other than 1 and the number itself.
So, it is a composite number
Posted by Gungun .. 4 years, 8 months ago
- 1 answers
Yogita Ingle 4 years, 8 months ago
Two digit odd positive numbers are 11,13,15,17...........99 are in A.P.
Here a = 11 and d = 2, tn= 99, n = ?
Sum of the n terms = (n/2)[2a+(n -1)d]
But tn = a + (n -1)d
⇒ 99 = 11+ (n-1)2
⇒ 99 -11 = (n-1)2
⇒ 88/2 = (n-1)
∴ n = 45.
subsitute n = 45 in sum of the n terms we obtain
⇒ s45 = (45/2)(2×11 + (45 -1)2)
⇒ s45 = (45/2)(110)
⇒ s45 = 45×55.
⇒ s45 = 2475.
∴sum of all two digit odd positive numbers = 2475.
Posted by Priyansha Pratap 4 years, 8 months ago
- 0 answers
Posted by Himanshu Verma 4 years, 8 months ago
- 1 answers
Gaurav Seth 4 years, 8 months ago
Nature of Roots
A quadratic equation ax2 + bx + c = 0 has
(i) two distinct real roots, if b2 – 4ac > 0,
(ii) two equal real roots, if b2 – 4ac = 0,
(iii) no real roots, if b2 – 4ac < 0.
Since b2 – 4ac determines whether the quadratic equation ax2 + bx + c = 0 has real roots or not, b2 – 4ac is called the discriminant of this quadratic equation
Posted by Alisha. Afreen 4 years, 8 months ago
- 5 answers
ಹರ್ಷನಂದ ಹರ್ಷನಂದ 4 years, 8 months ago
Posted by Pratham Rathod 4 years, 8 months ago
- 3 answers
Posted by Divya Mande 4 years, 8 months ago
- 3 answers
Posted by Mehul Kedia 4 years, 8 months ago
- 1 answers
Gaurav Seth 4 years, 8 months ago
Let four numbers in A.P. be a – 3d, a – d, a + d and a + 3d.
(a – 3d) + (a – d) + (a + d) + (a + 3d) = 32
⇒ 4a = 32
⇒ a = 8
Product of extremes = (a – 3d) (a + d) + (8 – 3d) (8 + 3d) = 64 – 9d2
Product of means = (a – d) (a + d) = (8 – d) (8 + d) = 64 – d2
Product of extremes: product of means = 7:15 (Given)
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When a = 8 and d = 2
a – 3d = 8 – 3 × 2 = 8 – 6 = 2
a – d = 8 – 2 = 6
a + d = 8 + 2 = 10
a + 3d = 8 + 3 × 2 = 8 + 6 = 14
When a = 8, d = –2
a – 3d = 8 – 3 × (–2) = 8 + 6 = 14
a – d = 8 – (–2) = 8 + 2 = 10
a + d = 8 – 2 = 6
a + 3d = 8 + 3 × (–2) = 8 – 6 = 2
∴ The four numbers are 2, 6, 10 and 14.
Posted by Mehul Kedia 4 years, 8 months ago
- 1 answers
Gaurav Seth 4 years, 8 months ago
Let the three terms be a-d,a,a+d
Sum=a-d+a+a+d=48
Solving this, we get
3a=48
a=16
a(a-d)-4(a+d)=12
Keeping the value of a in above equation
16(16-d)-4(16+d)=12
Solving this, we get
d=9
Therefore terms are 7,16,25 (ans)
or
Let the first three terms of the AP be (a-d), a and (a+d). Then,
Hence, the first three terms of the AP are 7, 16, and 25.
Posted by Mehul Kedia 4 years, 8 months ago
- 1 answers
Gaurav Seth 4 years, 8 months ago
The parts are in AP, let the terms be a-d, a and a+d
where the first term =a-d
the common difference =d
Given, sum =24cm
a-d + a+a +d = 24cm
3a = 24cm
a =8cm
Given: Product of the terms = 440
(a-d) (a) (a+d) = 440
(8-d) (8+d) (8) =440
(8-d)(8+d) = 55
82 -d2 = 55
64 - d2 = 55
d2 = 64-55
d2 = 9
d = 3 or -3
AP- a-d , a, a+d
so the AP is 8-3, 8, 8+3 or 8+3,8,8-3
AP = 5, 8, 11 or 11, 8, 5
Posted by Lomror ? 4 years, 8 months ago
- 1 answers
Gaurav Seth 4 years, 8 months ago
4 : 3
Let Sn and S'n be the sums of n terms of two AP’s and T11 and T11 be the respective 11th terms. Then
Now put n = 21. We get
Note: If ratio of sum of n terms of two APs is given in terms of n and ratio of their pth terms is to be found then put n = 2p - 1. Here we put n = 11 x 2 - 1 = 21.
Posted by Braj Yadav 4 years, 8 months ago
- 2 answers
Shiva Gowri 4 years, 8 months ago
Posted by Sumit Rathi 4 years, 8 months ago
- 1 answers
Gaurav Seth 4 years, 8 months ago
- he prime factors are: 2 x 2 x 2 x 3 x 3 x 5 x 7 x 13 x 17
- Written in exponential form: 23 x 32 x 51 x 71 x 131 x 171
| 556,920 | |||||||||
 |
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| 2 | 278,460 | ||||||||
|
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| 2 | 139,230 | ||||||||
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|||||||||
| 2 | 69,615 | ||||||||
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|||||||||
| 3 | 23,205 | ||||||||
|
|||||||||
| 3 | 7,735 | ||||||||
|
|||||||||
| 5 | 1,547 | ||||||||
|
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| 7 | 221 | ||||||||
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| 13 | 17 |
Posted by Samruddhi Br 4 years, 8 months ago
- 2 answers
Gaurav Seth 4 years, 8 months ago
In Fig. 12.25, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.
Solution:
Side of square = 14 cm
Four quadrants are included in the four sides of the square.
∴ Radius of the circles = 14/2 cm = 7 cm
Area of the square ABCD = 142 = 196 cm2
Area of the quadrant = (πR2)/4 cm2 = (22/7) ×72/4 cm2
= 77/2 cm2
Total area of the quadrant = 4×77/2 cm2 = 154cm2
Area of the shaded region = Area of the square ABCD – Area of the quadrant
= 196 cm2 – 154 cm2
= 42 cm2
Posted by Samruddhi Br 4 years, 8 months ago
- 1 answers
Gaurav Seth 4 years, 8 months ago
Solution:
ABC is an equilateral triangle.
∴ ∠ A = ∠ B = ∠ C = 60°
There are three sectors each making 60°.
Area of ΔABC = 17320.5 cm2
⇒ √3/4 ×(side)2 = 17320.5
⇒ (side)2 =17320.5×4/1.73205
⇒ (side)2 = 4×104
⇒ side = 200 cm
Radius of the circles = 200/2 cm = 100 cm
Area of the sector = (60°/360°)×π r2 cm2
= 1/6×3.14×(100)2 cm2
= 15700/3cm2
Area of 3 sectors = 3×15700/3 = 15700 cm2
Thus, area of the shaded region = Area of equilateral triangle ABC – Area of 3 sectors
= 17320.5-15700 cm2 = 1620.5 cm2
Posted by Uday Kumar 4 years, 8 months ago
- 1 answers
Yogita Ingle 4 years, 8 months ago
Let A(2,3), B(−1,0) and C(2,−4) are the vertices of △ABC
Area of triangle= 1/2 [x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]
Here (x1,y1)=(2,3)
(x2,y2)=(−1,0)
(x3,y3)=(2,−4)
= 1/2 [2(0+4)−1(−4−3)+2(3−0)]
= 1/2 [8+7+6]
= 21/2 sq.unit
Posted by Muthu Pradeepa Rajesh 4 years, 8 months ago
- 1 answers
Yogita Ingle 4 years, 8 months ago
Given points A(2,−5) and B(−2,9)
Let the points be P(x,0).
So, AP=PB and AP2=PB2
⇒(x−2)2+(0+5)2=(x+2)2+(0−9)2
⇒x2+4−4x+25=x2+4+4x+81
⇒x2+29−4x=x2+85+4x
⇒−4x−4x=85−29
⇒−8x=56
⇒x=−7
Hence, point on the x-axis which is equidistant from (2,−5) and (−2,9) is (−7,0).
Posted by Lomror ? 4 years, 8 months ago
- 2 answers
Yogita Ingle 4 years, 8 months ago
Let the numbers be x and 3x
Then,
(x+5)/ (3x+5) =21
⇒2x+10=3x+5
⇒x=5
Hence, the numbers are 5 and 15
∴ Sum of numbers = 20
Posted by Srimeenakshi Kr 4 years, 8 months ago
- 1 answers
Yogita Ingle 4 years, 8 months ago
Let an equilateral triangle Δ ABC : AB = BC = AC.
vertices are A(3,2) and B(-3,2) and C(x,y).
The mid point of the side AB is M (0,2).
(AB)² = (3+3)² + (2-2)² = 6² + 0² = 36
AB = 6 cm
AB = BC = AC = 6 unit.
AM = 3 unit.
As two vertices are A(-3,2) and B(3,0) so that Third vertex will be at
Y-axis(x=0).
so that
Third vertex C(0,y) and it is located below the origin.
Hence it contains the origin.
Now,
y² = (AC)² - (AM)²
⇒ y² = 6² - 3² = 36 - 9 =27
y = √27 = 3√3 unit
Hence Third vertices of the triangle are A(-3,2) , B(3,2) and C(0,3√3).
Posted by Alisha. Afreen 4 years, 8 months ago
- 0 answers
Posted by Alisha. Afreen 4 years, 8 months ago
- 1 answers
ಹರ್ಷನಂದ ಹರ್ಷನಂದ 4 years, 8 months ago
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Jiya Bhayana 4 years, 8 months ago
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