71. the equation `x^2/(16-lamda)+y^2/(9-lamda)`=1 represents a hyperbola when `a<lamda<b` then the value of `a+b` is
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the given equations arex2 + y2 = 16 …(i)and y2 = 6x …(ii)on solving the equation (i) and (ii),or x2 + 6x = 16or x2 + 6x – 16 = 0or (x + 8)(x – 2) = 0or x = 2 or – 8 is not possible solution∴ when x = 2, y = ± = ± = ±equation (i) is a circle with centre (0, 0) and meets axes at (±4,0), (0,±4)equation (ii) represents a parabola with axis as x - axis.points of intersection are a () and c (2,)these are shown in the graph below:area bounded by the circle and parabola= 2[area (oado) + area (adba)]area of circle = π(r) 2= π(4)2= 16π sq. unitsthus, required areathe area of the circle x2 + y2 = 16 which is exterior the parabola y2 = 6x is in the given expressionwe know that(a + b)2 = a2 + 2ab + b2(a-b)2 = a2-2ab + b2hence4[a2 + 2ab + b2] – 9[a2-2ab + b2]4a2 + 8ab + 4b2 - 9a2 + 18ab - 9b226ab – 5a2 - 5b225ab + ab – 5a2 – 5b2[25ab – 5a2] + [ab – 5b2]5a[5b – a] – b[5b – a][5a – b][5b – a]hence the factors 4 (a + b)2 – 9 (a – b)2 are [5a – b] and [5b – a] if the figure below represents a parabola, identity the physical&nbsp;quantities Definition used:  

Congruence.  

Let m be an integer greater than 1. If xand yare integers, then x is congruent to y modulo m if x — y is divisible by m. It can be represented as \(\displaystyle{x}={y}{\left({b}\ mod\ {m}\right)}\). This relation is called as congruence modulo m.  

Calculation:   

Consider the values and congruence classes modulo m,  

\(\displaystyle{\left[{13}\right]}^{{{6}}}\) and m = 15.  

It is evident that [x]=[y] if \(\displaystyle{\left[{x}={y}{\left({b}\ mod\ {m}\right)}\right]}\). Since \(\displaystyle{13}=-{2}{\left({b} \ mod \ {15}\right)}\).  

Then,  

[13]=[-2]  

\(\displaystyle{\left[{13}\right]}^{{{6}}}={\left[-{2}\right]}^{{{6}}}\)  

\(\displaystyle={\left[{\left(-{2}\right)}^{{{6}}}\right]}\)  

=[64]  

Since, by the principal or congruence \(\displaystyle{64}={4}{\left({b}\ mod \ {15}\right)}\). Therefore, \(\displaystyle{\left[{13}\right]}^{{{6}}}={\left[{4}\right]}\) in \(\displaystyle{Z}\_{{{15}}}\).  

Therefore, the value of the expression is [4]. ?solution 23cq


a) equation for loop abgefa is (35.5)i1 + (6..25 = 21 v
b) equation for loop cbgedc is ( 9.25i2 - ( 6.25i3 = +15 v let abc be the trianglewhere ab=(a-1)2 cmbc=2√a cmca=(a+1) cmab2=(a-1)2=a2+1-2abc2=(2√a)2=4a2ca2=(a+1)2=a2+1+2ahence ab2+bc2=ac2so Δabc is a right angles triangle at b express as a sum of perfect square 16 a square + b square + 9 c square minus 4 a b minus 3 b c - 12 ca? method a.increment() overlaps in class b.   
 therefore, regardless of the type of the variable obj, will be called the method corresponding to the actual type of the object - that is, b.increment().  
 
this method will use the variable i, declared in the same class b and having an initial value of 10.  
 
the resulting value - 11. given;let ⇒ p(k+1) is true when p(k) is true.∴ by mathematical induction is true for all natural number n≥2. Δl =l αΔt =2 •10⁻²•14•10⁻⁶•70=1.96•10⁻⁵m=1.96•10⁻² mm