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Q.No2
1) Let the length of other two sides be x
According to the Pythagoras theorem
(10)² = x² + x²
2x² = 100
x² = 50
x² + 0x - 50 = 0
x = ±√(50)
x = ±√(25×2)
x = ±5√2
x = -5√2 will be rejected becoz lenght can't be negative.
So, x = 5√2 cm
So, length of other two sides are 5√2cm and 5√2cm
2) Let the ist odd integer be = x
it's consecutive odd integer will be x + 2
ACCORDING TO THE GIVEN QUESTION
x² + ( x + 2 )² = 970
x² + x² + 4 + 4x = 970
2x² + 4x + 4 = 970
2x² + 4x = 970 - 4
2x² + 4x = 966
x² + 2x -483 = 0
x² + 23x - 21x - 483 = 0
x( x + 23 ) - 21 ( x + 23 ) = 0
( x - 21 ) ( x + 23 ) = 0
x = 21 OR x = -23
x = -23 will be rejected becoz in give. question it is clearly written that positive odd integer's.
So, x = 21
one odd integer is 21 And another odd integer is 21 + 2 I,e 23
1. In an isosceles triangle two sides are equal
If the sides of an isosceles triangle other than hypotenus are x and y
we can say that x=y
By Pythagoras property
x^2+x^2= 10^2
2x^2=100
2x^2+0x-100=0
x^2+0x-50=0
2. Let the two odd positive integers be x and x+2
x^2+(x+2)^2=970
x^2+x^2+4x+4=970
2x^2+4x+4-970=0
2x^2+4x-966=0
x^2+2x-483=0
3.
SP=96
Let CP be x
Profit =SP-CP
Profit=96-x
Profit%=(Profit/CP)*100
Profit%=((96-x)/x)*100
Given that
Profit%=CP
(96-x)/x*100=x
100=x*x/ (96-x)
100=x^2/(96-x)
100 (96-x)=x^2
9600-100x=x^2
x^2+100x-9600=0
Hope it helps......