Math, asked by vaniviji0653, 1 month ago

1.The value of tan A is always less than 1. 2. cos A is the abbreviation used for the cosecant of angle A.
3. The value of cos  increases as  increases.
4. Zeroes of quadratic polynomial x 2 + 7x + 10 are 2 and -5
5. Sum of zeroes of 2x 2 - 8x + 6 is -4
6. A polynomial of degree n has exactly n zeros.
7. Graph of a quadratic polynomial is an ellipse.
8. 1/0 is not rational.
9. Every fraction is a rational number.​

Answers

Answered by ShashikantKumarGupta
1

Answer:

∠B=90

∠B=90 o

∠B=90 o , AB=3,BC=4,AC=5

∠B=90 o , AB=3,BC=4,AC=5tanA=

∠B=90 o , AB=3,BC=4,AC=5tanA= 3

∠B=90 o , AB=3,BC=4,AC=5tanA= 34

∠B=90 o , AB=3,BC=4,AC=5tanA= 34

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) True

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2 (12x)

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2 (12x) 2

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2 (12x) 2 =(5x)

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2 (12x) 2 =(5x) 2

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2 (12x) 2 =(5x) 2 +BC

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2 (12x) 2 =(5x) 2 +BC 2

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2 (12x) 2 =(5x) 2 +BC 2 ,

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2 (12x) 2 =(5x) 2 +BC 2 , BC=

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2 (12x) 2 =(5x) 2 +BC 2 , BC= 119x

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2 (12x) 2 =(5x) 2 +BC 2 , BC= 119x

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2 (12x) 2 =(5x) 2 +BC 2 , BC= 119x

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2 (12x) 2 =(5x) 2 +BC 2 , BC= 119x (it is possible) Pythagoras theorem

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2 (12x) 2 =(5x) 2 +BC 2 , BC= 119x (it is possible) Pythagoras theorem(iii) False

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2 (12x) 2 =(5x) 2 +BC 2 , BC= 119x (it is possible) Pythagoras theorem(iii) FalsecosA is cosine A, cscA is cosecant

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2 (12x) 2 =(5x) 2 +BC 2 , BC= 119x (it is possible) Pythagoras theorem(iii) FalsecosA is cosine A, cscA is cosecant (iv) False

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2 (12x) 2 =(5x) 2 +BC 2 , BC= 119x (it is possible) Pythagoras theorem(iii) FalsecosA is cosine A, cscA is cosecant (iv) FalsecotA = cotangent of ∠A not product of cot and A.

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2 (12x) 2 =(5x) 2 +BC 2 , BC= 119x (it is possible) Pythagoras theorem(iii) FalsecosA is cosine A, cscA is cosecant (iv) FalsecotA = cotangent of ∠A not product of cot and A.(v) False

∠B=90 o , AB=3,BC=4,AC=5tanA= 34 >d. ...(AC 2 =AB 2 +BC 2 )(ii) TrueAC 2 =AB 2 +BC 2 (12x) 2 =(5x) 2 +BC 2 , BC= 119x (it is possible) Pythagoras theorem(iii) FalsecosA is cosine A, cscA is cosecant (iv) FalsecotA = cotangent of ∠A not product of cot and A.(v) Falsesin θ=

θ= hypotenuse

θ= hypotenuseperpendicular

θ= hypotenuseperpendicular

θ= hypotenuseperpendicular

θ= hypotenuseperpendicular sinθ>1;

θ= hypotenuseperpendicular sinθ>1; 3

θ= hypotenuseperpendicular sinθ>1; 34

θ= hypotenuseperpendicular sinθ>1; 34

θ= hypotenuseperpendicular sinθ>1; 34 >1 [∵ hypotenuse > base > perpendicular]

θ= hypotenuseperpendicular sinθ>1; 34 >1 [∵ hypotenuse > base > perpendicular]sinθ will always less than 1.

Answered by omkarpatil05
0

Answer:

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