please answer question 5 and 6
tomorrow is my math exam please please please please please please please give
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6.
we have to divide 29 into two parts so,
x + y = 29
x = 29-y
then
x^2 + y^2 = 425
(29 - y )^2 + y^2 = 425
841 + y^2 + 58y + y^2 = 425
2y^2 + 58y + 416 = 0
y^2 + 29y + 208 = 0
y^2 + 13y +16y + 208 = 0
y(y+13) + 16(y+13) = 0
(y+16)(y+13) = 0
y = -13 and y = -16
then,
x = 42 and x = 45
therefore there are two ways to divide 29 according to above circumstances
those are 29 = 42 - 13 and 45 - 16.
we have to divide 29 into two parts so,
x + y = 29
x = 29-y
then
x^2 + y^2 = 425
(29 - y )^2 + y^2 = 425
841 + y^2 + 58y + y^2 = 425
2y^2 + 58y + 416 = 0
y^2 + 29y + 208 = 0
y^2 + 13y +16y + 208 = 0
y(y+13) + 16(y+13) = 0
(y+16)(y+13) = 0
y = -13 and y = -16
then,
x = 42 and x = 45
therefore there are two ways to divide 29 according to above circumstances
those are 29 = 42 - 13 and 45 - 16.
khushboo18:
what do u want to say
Answered by
0
Solve this equation to get the solution. +ve root will be the answer
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