Vitamin B3 is also known as

1 Answer

Answer :

 Ans. Pantathenic acid

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Description : Corneal vascularisation is found in deficiency of the vitamin: (A) B1 (B) B2 (C) B3 (D) B6

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Description : What is true of nicotinic acid as well as nicotinamide: A. Both possess vitamin B3 activity B. Both cause cutaneous vasodilatation C. Both lower plasma triglyceride and VLDL levels D. Both cause hyperglycaemia after prolonged medication

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Description : If a + b + c =0, then a3+b3 + c3 is equal to -Maths 9th

Last Answer : (d) Now, a3+b3 + c3 = (a+ b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc [using identity, a3+b3 + c3 – 3 abc = (a + b + c)(a2+b2+c2 –ab–bc-ca)] = 0 + 3abc [∴ a + b + c = 0] a3+b3 + c3 = 3abc.

Description : If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3 – 3abc = -25. -Maths 9th

Last Answer : Prove that a3 +b3 +c3 – 3abc = -25

Description : Prove that (a +b +c)3 -a3 -b3 – c3 =3(a +b)(b +c)(c +a). -Maths 9th

Last Answer : Solution to (a +b +c) 3 -a 3 -b 3 – c 3 =3(a +b)(b +c)(c +a)

Description : If a + b + c =0, then a3+b3 + c3 is equal to -Maths 9th

Last Answer : (d) Now, a3+b3 + c3 = (a+ b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc [using identity, a3+b3 + c3 – 3 abc = (a + b + c)(a2+b2+c2 –ab–bc-ca)] = 0 + 3abc [∴ a + b + c = 0] a3+b3 + c3 = 3abc.

Description : If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3 – 3abc = -25. -Maths 9th

Last Answer : Prove that a3 +b3 +c3 – 3abc = -25

Description : Prove that (a +b +c)3 -a3 -b3 – c3 =3(a +b)(b +c)(c +a). -Maths 9th

Last Answer : Solution to (a +b +c) 3 -a 3 -b 3 – c 3 =3(a +b)(b +c)(c +a)

Description : The product (a + b) (a – b) (a2 – ab + b2) (a2 + ab + b2) is equal to (a) a6 + b6 (b) a6 – b6 (c) a3 – b3 (d) a3 + b3 -Maths 9th

Last Answer : answer:

Description : If a + b + c = 9 and ab + bc + ca = 23, then a3 + b3 + c3 – 3 abc = (a) 108 (b) 207 (c) 669 (d) 729 -Maths 9th

Last Answer : a+b+c=9 and a2+b2+c2=35 Using formula, (a+b+c)2=a2+b2+c2+2(ab+bc+ca) 92=35+2(ab+bc+ca) 2(ab+bc+ca)=81−35=46 (ab+bc+ca)=23 using formula, (a3+b3+c3)−3abc=(a2+b2+c2−ab−bc−ca)(a+b+c) a3+b3+c3−3abc=(35−23)×9=9×12=108

Description : If a – b = 4 and ab = 21, find the value of a3-b3. -Maths 9th

Last Answer : Given, a−b=4,ab=21 Then, ⇒(a−b)3=a3−3a2b+3ab2−b3=a3−3ab(a−b)−b3 ⇒43=a3−b3−3(21)(4) ⇒64+252=a3−b3 ∴a3−b3=316

Description : If a + b = 10 and ab = 21, find the value of a3 + b3. -Maths 9th

Last Answer : Given, a+b=10,ab=21 then, ⇒(a+b)3=a3+3ab(a+b)+b3 ⇒(10)3=a3+b3+3(21)(10) ⇒1000−630=a3+b3 ∴a3+b3=370

Description : If the roots ff the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 are equal, then prove that either a = 0 or a3 + b3 + c3 = 3abc -Maths 10th

Last Answer : (c2 – ab) x2 + 2(bc - a2 ) x+ (b2 – ac) = 0 Comparing with Ax2 + Bx + C = 0 A = (c2 – ab), B = 2(bc - a2 ) and C = b2 – ac According to the question, B2 - 4AC = 0 Put the values in the above equation we get 4a(a3 + b3 + c3 -3abc) = 0 hence, a = 0 or a3 + b3 + c3 = 3ab

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Description : hydraulically equivalent, the relationship which holds good, is A. D8/3 = 4 b8/3 B. D3/8 = 4 b3/8 C. D2/3 = 4 b2/3 D. D3/2 = 4 b3/2

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Description : If the centroid of △ABC in which A(a,b), B(b,c) and C(c,a) is at the origin, then the value of a3 + b3 + c3 is: (a) abc (b) 2abc (c) 3abc (d) 0

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