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  • 2 answers
Given : Half the perimeter of a garden whose length is 4 m more than its width is 36 m. To do: We have to find the dimensions of the garden. Solution : Perimeter of a rectangle of length l and breadth b=2(l+b). Let the length of the rectangle be l and the width of the rectangle be b. Therefore, Half the perimeter of rectangle = 2 (l + b) 2 Half the perimeter of rectangle =l+b. Given that half the perimeter of the rectangular garden =36 m. l+b=36 m......(i) It is given that, length of the rectangular garden is 4 m more than its width. This implies, l=b+4.......(ii) Substituting (ii) in (i), we get, b+4+b=36 2b=36−4 2b=32 b= 32/2 b=16 m Substitute b=16 in (ii) l=16+4 l=20 m Therefore, the length of the rectangular garden is 20 m and the width of the rectangular garden is 16 m.

Ankit Class 2 years ago

Let length be l and breadth be b of rectangular garden. l= b+4. (l+b)/2 = 36 . l+b= 72. b+4 + b = 72. 2b+4= 72. 2b= 68. b= 34 m. l=b+4 = 34+4= 38 m.
  • 1 answers

Ankit Class 2 years ago

a=121 , d= -4 let l=o . a+ (n-1)d = l . 121+(n-1)x-4= 0. (n-1)=121/4. n= 30.25+1. n = 31.25 , Here we know that n always be natural number hence n should be 31. n= 31.
  • 1 answers

_Jass_ Mahey_ 2 years ago

Let 'a' be a postive odd integer also. Let 'q' be quotient and 'r' be the remainder after dividing by six. Then a=6q+r, where 0<r<6. Postive integer are 1,2,3,4,5 a=6q+0, 6q+1, 6q+2, 6q+3, 6q+4, 6q+5. But a=6q, 6q+2 and 6q+4 are even. When a is add it is the form 6q+1, 6q+3, 6q+5 for some integer q. .°. Hence proved...
  • 1 answers

Ankit Class 2 years ago

(2/a )=(3/a+p)= (7/28). 2/a= 7/28 . 2/a= 1/4 . a= 8. (3/p+8)= 7/28. (3/p+8)= 1/4. P+8 = 12. P= 4 .
  • 2 answers

Ankit Class 2 years ago

0° , 30° ,45° ,60° , 90° ,120° , 135° ,150° , 180° any many more

Ankit Class 2 years ago

😂
  • 1 answers

Ankit Class 2 years ago

Subtract 1,2,3 from these no we get 1250, 9375, 15625 now use Euclid algorithm first in any of two then you get HCF of those two number then use again Euclid algorithm in HCF that you got with the third number
  • 2 answers

Prince Sharma 2 years ago

2sinQ=1 =>sinQ=1/2 =>sinQ=sin30° [On comparing] =>Q=30°
Sin30°
  • 3 answers

Prince Sharma 2 years ago

(4 -1)² +(p - 0)² = 5². => 9 + p² =25. => p² = 16. => p = 4
4

_Jass_ Mahey_ 2 years ago

P=3
  • 3 answers

Ankit Class 2 years ago

D=√ (-11+2/3)^2 +(5-5)^2 . D= √ 9 . D= 3 unit

Ankit Class 2 years ago

I don' t like much calculation because l am little weak in calculation 🤣

Ankit Class 2 years ago

Use distance formula
  • 2 answers

Ankit Class 2 years ago

Ratio of any two side in a right angle triangle whose specific name was kept by their specific ratio like p/ h , b/ h , p/ b etc
Sin - Sine Cos - Cosine Tan - Tangent Cosec - Cosecant Sec - Secant Cot - Cotangent
  • 2 answers

Maya Shukla 2 years ago

x=2 and -2/5

Prince Sharma 2 years ago

5x²-8x-4=0 => 5x²-10x+2x-4=0 => 5x(x-2) + 2(x-2)=0 => (x-2)(5x+2)=0 => x=2,-2/5
  • 2 answers

Ankit Class 2 years ago

HCF (A,B) = 1 . Because you can check by taking any natural number in place of n .

Prem Latha 2 years ago

Ans
  • 2 answers

Maya Shukla 2 years ago

Question acha hai 🤟

Maya Shukla 2 years ago

660- 2,2,3,5,11. 704-2,2,2,2,2,2,11. Answer hai 22
  • 5 answers

Preeti Dabral 2 years ago

Let distance between the two towers = AB = x m
and height of the other tower = PA = h m
Given that, height of the tower = QB = 30 m and QAB = 60°, PBA = 30°.

Now, in ΔQAB,tan60=QBAB=30x
3=30x
x=x23x3x33x+0.5π
and in ΔPBA
tan30=PAAB=hx
13=h103 [x=103m]
h=10m
Hence, the requied distance and height are 103 and 10m, respectively.

Ankit Class 2 years ago

Ye figure draw kaise ki ho batao

Ankit Class 2 years ago

PQ is the length of the wire attached to the tops of both the towers

Ankit Class 2 years ago

PQ=√x^2 +(30-h)^2 . PQ=√(10√3)^2 + (30-10)^2 . PQ=√ 300+400. PQ=√700. PQ= 10√7 m

Ankit Class 2 years ago

Math standard mein ye question puchha tha n mere mein bhi puchha tha set 1 tha mera 21 feb ko paper tha n
  • 1 answers

Preeti Dabral 2 years ago

h = 20t - 16t²

= (20 x 3/2) - 16 x (3/2)²

= 30 - ( 16 x 9/4)

= 30 - 36 = -6

  • 1 answers

Preeti Dabral 2 years ago

The distance between the two points is 2√10 units .

  • 1 answers

Y Music 2 years ago

The distance between the points (0, 2√5) and (-2 √5, 0) is
  • 2 answers

_Jass_ Mahey_ 2 years ago

√16

Sahil Singh 2 years ago

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  • 5 answers

Maya Shukla 2 years ago

Answer hai -10

_Jass_ Mahey_ 2 years ago

By joining B and D. You will get two triangles AND and BCD. Now the area of ∆ABD = 1/2 (-5(-5-5)+(-4) (5-7) +4(7+5) = 1/2(50+8+48) =106/2=53square units Also, the area of ∆BCD=1/2(-4(-6-5) -1(5+5) +4(-5+6) =1/2(44-10+4) =19 square units So, the area of quadrilateral ABCD=53+19=72square units [Note:to find thr area of a polygon, we divide it into triangular regions, which have no common area of these regions. ]

Priyom Shah 2 years ago

-10

Sahil Singh 2 years ago

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Vettri Vel 2 years ago

Don't know that is the answer
  • 2 answers

Rishika Kabra 2 years ago

Sin²a+ cos²a)²= (sin²a)²+ (cos²a)²+2 sin²a cos²a=1 Sin⁴a+cos⁴a= 1-2sin²a cos²a 1-2 sin²a cos²a/1-2sin²a cos²a=1 1=1 LHS= RHS hence proved

Rishika Kabra 2 years ago

(Sin²a+ cos²a)²= (sin²a)²+ (cos²a)²+2 sin²a cos²a
  • 1 answers

Ankit Class 2 years ago

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  • 1 answers

Preeti Dabral 2 years ago

Let’s consider the number of 20 rupee notes to be x
And the number of 5 rupee notes be y

Then, according to the given conditions, we have

20x+5y=380     … (i) and

5x+20y=380−60

⇒5x+20y=320     … (ii)

Now, multiplying eq(i) by 4 and subtracting with eq(ii), we get

(80x+20y)−(5x+20y)=1520−320

⇒75x=1200

x=120075x=16

On substituting the value of x in eq(i), we get

20(16)+5y=380

⇒320+5y=380

⇒5y=380−320=60

y=605y=12

Therefore, 

The number of 20 rupee notes =16 and 

The number of 5 rupee notes =12

  • 1 answers

Preeti Dabral 2 years ago


Given A circle with centre O and an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact.
To Prove: PTQ = 2OPQ
Proof: Let PTQ = θ
Since TP, TQ are tangents drawn from point T to the circle.
TP = TQ
 TPQ is an isoscles triangle
 TPQ = TQP = 12 (180o - θ) = 90o - \fracθ2
Since, TP is a tangent to the circle at point of contact P
\therefore \angleOPT = 90o
\therefore \angleOPQ = \angleOPT - \angleTPQ = 90o - (90o\frac12 \theta) = \fracθ2= \frac12\anglePTQ
Thus, \anglePTQ = 2\angleOPQ

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