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  • 1 answers

Saurabdeep Kaur 2 years, 1 month ago

Ans: 51
4+8
  • 5 answers

Vandana Yadav 2 years, 1 month ago

12

Abhishek Mishra 2 years, 1 month ago

12

Mamta Nayak 2 years, 1 month ago

12

Sk Siddiq 2 years, 1 month ago

Answer is 12

Rubul Pathori Pathori 2 years, 1 month ago

12
  • 2 answers

Shristi Chauhan 2 years, 1 month ago

Answer is large

Sk Siddiq 2 years, 1 month ago

Answer is large
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Dipika Rautaray 2 years, 1 month ago

Thank you for your answer but i want the value of the equation

Mahima Mahore 2 years, 1 month ago

Polynomial
  • 1 answers

Manti Kumari Bhogta 2 years, 1 month ago

दिल्ली स युग डीएफ उफ़ एक्स
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  • 2 answers

Saurabdeep Kaur 2 years, 1 month ago

X²+11x+9

Mamta Nayak 2 years, 1 month ago

(x+2)(x+9) x(x+2)9(x+1) (x+9)(x+2)
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Pranjal Singh 2 years ago

15876

Vansh Bhati 2 years, 1 month ago

Idk
  • 1 answers

Ritica Singh 2 years, 1 month ago

pairs of numbers that do not have any common factor other than 1. Eg: 2,21 Factor of 2: 1,2 Factor of 21:1,3,7,21 Common factor:1
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Shanu Pratap 2 years, 1 month ago

First let's find out semi perimeter of this triangle which is 21 By using heron's formula ( s√s(s-a)(s-b)(s-c) u have to calculate it's area then by using 1/2 *b*h u can get it's height or altitude
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Shanu Pratap 2 years, 1 month ago

Remainder theorem: If a polynomial P(x) is divided by (x-c), then the remainder is equal to P(c). The given polynomial are p(t)=4t3−st2+7p(t)=4t3−st2+7 q(t)=t2+st+8q(t)=t2+st+8 Using remainder theorem the remainder of p(t)t−1p(t)t−1is p(1) and the remainder ofq(t)t−1q(t)t−1 is q(1). Substitute t=1 in the given functions. p(1)=4(1)3−s(1)2+7⇒4−s+7=11−sp(1)=4(1)3−s(1)2+7⇒4−s+7=11−s q(1)=(1)2+s(1)+8=1+s+8=9+sq(1)=(1)2+s(1)+8=1+s+8=9+s It is given that if p(t) and q(t) divided by (t-1), then the remainder is same. p(1)=q(1) Substitute these values. 11-s=9+s Add s on both sides. 11=9+s+s 11=9+2s Subtract 9 from both sides. 11-9=2s 2=2s Divide both sides by 2. 1=s
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Shanu Pratap 2 years, 1 month ago

Let ABC be the △ with sides BC,AC and AB opposite to the angles A,B and C being in length equal to a,b and c respectively. To prove:- AL+BM+CN<AB+BC+CA Construction:- Draw perpendiculars AL,BM and CN from A,B and C to opposite sides meeting respectively. Now, In △ACL, Using pythagoras theorem, AC2=AL2+CL2 ⇒AC2>AL2 ⇒AC>AL.....(1) Similarly, AB>BM.....(2) BC>CN.....(3) Adding equation (1),(2)&(3), we get] AB+BC+AC>AL+BM+CN Hence proved.
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Abhijeet Anand 2 years, 1 month ago

Triangle Solve for area A≈43.82cm² a Side -- 8cm b Base --11cm Perimeter -- 32cm
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Achyutha Bs 2 years, 1 month ago

Supplement=sum of two angles=180° Let two angles be x and y x+y=180° here x=y x+x=180° 2x=180° x=180°/2 =90° Therefore angles i.e. equal to it's supplement is 90°
  • 2 answers

Achyutha Bs 2 years, 1 month ago

Axiom is a universal truth accepted by all,but it doesn't have any proof.

Bibhas Ranjan Nayak 2 years, 1 month ago

assumptions which are not related to geometry or particular to geometry

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