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Ask QuestionPosted by Pooja Fojdar 4 years, 9 months ago
- 5 answers
Posted by Kunal Verma 4 years, 9 months ago
- 2 answers
Kap Sharma 4 years, 9 months ago
Posted by Kusum Garg 4 years, 9 months ago
- 0 answers
Posted by Shivam Yadav 4 years, 9 months ago
- 2 answers
Yogita Ingle 4 years, 9 months ago
27,24,21....
d=24-27=-3 a=27
let nth term=0
a+(n-1)d=0
27+(n-1)(-3) =0
(n-1)(-3)=-27
n-1=9
n=10
Posted by Clark Marak 4 years, 9 months ago
- 0 answers
Posted by Harshit Pandey 4 years, 9 months ago
- 1 answers
Keshav Yadav 4 years, 9 months ago
Posted by Angad Deep 4 years, 9 months ago
- 2 answers
Posted by Satyam Kumar Yadav Kumar Yadav 4 years, 9 months ago
- 1 answers
Yogita Ingle 4 years, 9 months ago
Let ΔABC and ΔPQR be two similar triangles. AD and PM are the medians of ΔABC and ΔPQR respectively.
Posted by Dimpi 😦 4 years, 9 months ago
- 1 answers
Posted by Sneha Markam 4 years, 9 months ago
- 2 answers
Posted by Rakesh Kumar 4 years, 9 months ago
- 0 answers
Posted by Alisha. Afreen 4 years, 9 months ago
- 2 answers
Posted by Alisha. Afreen 4 years, 9 months ago
- 0 answers
Posted by Alisha. Afreen 4 years, 9 months ago
- 2 answers
Gaurav Seth 4 years, 9 months ago
Width of the track = 10 m
Distance between two parallel lines = 60 m
Length of parallel tracks = 106 m
DE = CF = 60 m
Radius of inner semicircle, r = OD = O’C
= 60/2 m = 30 m
Radius of outer semicircle, R = OA = O’B
= 30+10 m = 40 m
Also, AB = CD = EF = GH = 106 m
Distance around the track along its inner edge = CD+EF+2×(Circumference of inner semicircle)
= 106+106+(2×πr) m = 212+(2×22/7×30) m
= 212+1320/7 m = 2804/7 m
Area of the track = Area of ABCD + Area EFGH + 2 × (area of outer semicircle) – 2 × (area of inner semicircle)
= (AB×CD)+(EF×GH)+2×(πr2/2) -2×(πR2/2) m2
= (106×10)+(106×10)+2×π/2(r2-R2) m2
= 2120+22/7×70×10 m2
= 4320 m2
Posted by Nitesh Ganjhu 4 years, 9 months ago
- 2 answers
Rajneesh Payal 4 years, 9 months ago
Yogita Ingle 4 years, 9 months ago
Rewrite the equation as:
(x + 1)(x + 4)(x + 3)(x + 2) = 120
Multiply the first 2 and last 2 expressions:
{tex}(x^2 + 5x + 4)(x^2 + 5x + 6) = 120{/tex} ..... (1)
Let {tex}x^2 + 5x = y{/tex}
(1) becomes,
(y + 4)(y + 6) = 120
y2 + 10x + 24 = 120
y2 + 10x - 96 = 0
y2 + 16x - 6y - 96 = 0
y(y + 16) - 6(y + 16) = 0
(y + 16)(y - 6) = 0
y = -16 or 6
y2+ 5x = 16 or 6
When x2 + 5x = 6
x2 + 5x - 6 = 0
x2 + 6x - x - 6 = 0
x(x + 6) - 1(x + 6) = 0
(x + 6)(x - 1) = 0
x = -6, 1
When x^2 + 5x = -16
x^2 + 5x + 16 = 0
Using quadratic formula:
{tex}x = [-5 +/- sqrt (25 - 64)]/2{/tex}
The solutions are, -6, 1, [-5 - sqrt (25 - 64)]/2 and [-5 + sqrt (25 - 64)]/2
These are all the possible solutions. But the only real solutions are -6 and 1.
Posted by Madhaan Ss 4 years, 9 months ago
- 1 answers
Posted by Simranpreet Kaur 4 years, 9 months ago
- 1 answers
Yogita Ingle 4 years, 9 months ago
Given : Line segment joining the points (6,4)and(1,-7) is divided by x axis.
To find : The ratio in the line segment and the coordinates of point of division?
Solution :
Let the line segment points A=(6,4) and B=(1,-7)
Let the line segment divide by x-axis with point P=(x,0)
Let the ratio in which line segment divide be m:n=k : 1
Applying section formula,
Substitute the values,
Compare the y-coordinate,
So, The ratio is 4 :7.
Compare the x-coordinate,
Put the value of k,
So, The coordinate of point of division is
Posted by Vedika Roongta 4 years, 9 months ago
- 2 answers
Yogita Ingle 4 years, 9 months ago
Given:- A circle with center O,PA and PB are tangents drawn at ends A and B on chord AB.
To prove:- ∠PAB=∠PBA
Construction:- Join OA and OB
Proof:- In △AOB, we have
OA=OB(Radii of the same circle)
∠OAB=∠OBA.....(1)(Angles opposite to equal sides)
∠OAP=∠OBP=90(∵Radius⊥Tangent)
⇒∠OAB+∠PAB=∠OBA+∠PBA
⇒∠OAB+∠PAB=∠OAB+∠PBA(From (1))
⇒∠PAB=∠PBA
Hence proved.
Posted by Chinnu S 4 years, 9 months ago
- 1 answers
Gaurav Seth 4 years, 9 months ago
Polynomial P(x) = x² - 6 x + a
Given α and β are the roots. To find a , if 3 α + 2 β = 20. ---(1)
From the quadratic expression:
α + β = 6 ---(2)
and α β = a --- (3)
Multiply equation (2) by 2 and subtract from (1) to get:
α = 20 -12 = 8
Substitute this value in(2) to get:
β = 6-8 = -2
Substitute these in (3) to get: a = α β = -16.
Posted by Anu Budhwar 4 years, 9 months ago
- 1 answers
Gaurav Seth 4 years, 9 months ago
Let the tens place digit be x.
And the units place digit be y.
According to the Question,
⇒ xy = 18
⇒ y = 18/x .....(i)
And, (10x + y) - 63 = 10y + x
⇒ 9x - 9y = 63
⇒ x - y = 7 .... (ii)
Putting y's value in Eq (ii), we get
⇒ x - 18/x = 7
⇒ x² - 18 = 7x
⇒ x² - 7x - 18 = 0
⇒ x² - 9x + 2x - 18 = 0
⇒x(x - 9) + 2(x - 9) = 0
⇒ (x - 9) (x + 2) = 0
⇒ x - 9 = 0 or x + 2 = 0
⇒ x = 9, - 2 (As x can't be negative)
⇒ x = 9
Putting x's value in Eq (i), we get
⇒ xy = 18
⇒ 9y = 18
⇒ y = 18/9
⇒ y = 2
Number = 92
Hence, the required number is 92.
Posted by Kundan Mandal 4 years, 9 months ago
- 0 answers
Posted by Vivek Pandey 4 years, 9 months ago
- 2 answers
Yogita Ingle 4 years, 9 months ago
Let the length and breadth of the park be l and b.
Perimeter = 2 (l + b) = 80
l + b = 40
Or, b = 40 - l
Area = l×b = l(40 - l) = 40l - l240l - l2 = 400
l2 - 40l + 400 = 0
Comparing this equation with al2 + bl + c = 0, we get
a = 1, b = -40, c = 400
Discriminant = b2 - 4ac
(-40)2 - 4 × 400
= 1600 - 1600 = 0
b2 - 4ac = 0
Therefore, this equation has equal real roots. And hence, this situation is possible.
Root of this equation,l = -b/2a
l = (40)/2(1) = 40/2 = 20
Therefore, length of park, l = 20 m
And breadth of park, b = 40 - l = 40 - 20 = 20 m.
Posted by Chanchala Kumari 4 years, 9 months ago
- 0 answers
Posted by Anand Kumar 4 years, 9 months ago
- 2 answers
Yogita Ingle 4 years, 9 months ago
sin2A=2sinA
2sinA.cosA-2sinA =0
2sinA(cosA-1)=0
either 2sinA=0
sinA=0
sinA=sin0°
A= 0°
or cosA -1=0
cosA=1
cosA=cos 0°
A=0° , Answer
Posted by Bhumika R 4 years, 9 months ago
- 1 answers
Posted by Supriya Mishra 4 years, 9 months ago
- 0 answers
Posted by Shravan Verma 4 years, 9 months ago
- 1 answers
Yogita Ingle 4 years, 9 months ago
Odd number always end in one of the digits
1, 3, 5, 7, 9
Posted by Bhaskar Gusain 4 years, 9 months ago
- 2 answers
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Seenu B 4 years, 9 months ago
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