@Aman.Malhotra said:x^3+y^3 -xy
x=1/2,y=1/2
x+y=1??
@YouMadFellow said:
total petrol used = 55-5=50
mileage = 400/50 = 8km/ltr
for minimum cost, she should complete each junction with 0 ltrs
she already has 5 ltrs, so distance till first pump = 150-40 = 110km
between 2nd and 3rd = 50km
between 3rd and 4th = 200km
so price = 110/8*45+50/8*40+200/8*35 = 1743.75
mileage = 400/50 = 8km/ltr
for minimum cost, she should complete each junction with 0 ltrs
she already has 5 ltrs, so distance till first pump = 150-40 = 110km
between 2nd and 3rd = 50km
between 3rd and 4th = 200km
so price = 110/8*45+50/8*40+200/8*35 = 1743.75
@gautam22 said:but y r u putting x=y to simplify the quation.....y r u doing so?......there shud be some reason behind putting that
for more insight into this matter...plzz consult @krum sir (demi - god)....this is my weak topic, i can't help much..
@priyalli said:A man borrows 6000at 5 % interest on reducing balance ,at the start of the year .If he pays 1200 at the end of each ear ,find the amount of loan outstanding at the beginning of the third year ???please post the detailed solution for this ques
4155?
@YouMadFellow said:
mileage=400/50=8
in 5 l she can go 40kms
for 110 km
she has to purchase
110/8 *45
second purchase=50/8*40
third purchase=200/8*35
4950+2000+7000 / 8
13950/8=1743.75=1744
@priyalli said:A man borrows 6000at 5 % interest on reducing balance ,at the start of the year .If he pays 1200 at the end of each ear ,find the amount of loan outstanding at the beginning of the third year ???please post the detailed solution for this ques
6000*5/100=300
6000+300-1200=5100
5100*5/100=255
5100+255-1200=4155??
@busar005 said:for more insight into this matter...plzz consult @krum sir (demi - god)....this is my weak topic, i can't help much..
@gautam22 said:@krum sir thoda dekhna
i solve questions on whim, such explanations are beyond me, kahe ka sir?
chill sir aur ATDH sir ko tag kardo yar
chill sir aur ATDH sir ko tag kardo yar
@priyalli said:A man borrows 6000at 5 % interest on reducing balance ,at the start of the year .If he pays 1200 at the end of each ear ,find the amount of loan outstanding at the beginning of the third year ???please post the detailed solution for this ques
4155...?
a, b and c are the sides of a triangle. Equations ax^2 + bx + c = 0 and 3x^2 + 4x + 5 = 0 have a common root. Then angle C is equal to??
@Aman.Malhotra said:a, b and c are the sides of a triangle. Equations ax^2 + bx + c = 0 and 3x^2 + 4x + 5 = 0 have a common root. Then angle C is equal to??
3x^2+4x+5=0
d=-44
this has imaginary roots
@Aman.Malhotra said:a, b and c are the sides of a triangle. Equations ax^2 + bx + c = 0 and 3x^2 + 4x + 5 = 0 have a common root. Then angle C is equal to??
imaginary roots aa rahe hai...so triangle not possible?
@Aman.Malhotra said:a, b and c are the sides of a triangle. Equations ax^2 + bx + c = 0 and 3x^2 + 4x + 5 = 0 have a common root. Then angle C is equal to??
90 degrees?
2nd eqn has imaginary roots
Since imaginary roots occur in conjugates hence other root will also be the root of the 1st eqn
In essence both eqns r same
So cosC=9+16-25/2*3*4=0
Hence angle C=90
2nd eqn has imaginary roots
Since imaginary roots occur in conjugates hence other root will also be the root of the 1st eqn
In essence both eqns r same
So cosC=9+16-25/2*3*4=0
Hence angle C=90
@Aman.Malhotra said:a, b and c are the sides of a triangle. Equations ax^2 + bx + c = 0 and 3x^2 + 4x + 5 = 0 have a common root. Then angle C is equal to??
If both the roots are common then it is 90...
for one root common, I am not getting it..
for one root common, I am not getting it..
@Aman.Malhotra said:a, b and c are the sides of a triangle. Equations ax^2 + bx + c = 0 and 3x^2 + 4x + 5 = 0 have a common root. Then angle C is equal to??
roots are imaginary.
And imaginary roots occur in pairs
so has to be 90 degree
@Aman.Malhotra said:a, b and c are the sides of a triangle. Equations ax^2 + bx + c = 0 and 3x^2 + 4x + 5 = 0 have a common root. Then angle C is equal to??
90?
@maddy2807 said:roots are imaginary.And imaginary roots occur in pairsso has to be 90 degree
yeh konsa sa funda hai....??(imaginary roots occur in pair so ...90 deg?)
plz explain
@mailtoankit said:yeh konsa sa funda hai....??(imaginary roots occur in pair so ...90 deg?)plz explain
bhai.. in that case a b c will be 3,4,5 respectively...that will make abc a right angle triangle at C..
A person lent out some money for 1year at 6% per annum simple interest and after 18months he again lent out the same money at a simple interest of 24% per annum. In both cases he got Rs.4704. What was the amount that he lent out if interest is paid half yearly?a rs 4000 b rs 4200 c rs 4400 d rs 3600
The L.C.M of 5/2, 8/9, 11/14 ???