The equation whose line is parallel to x-axis is and passing through the point (2, 1/3) is
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Answer:
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3=a(1)+b(−1)+c
3=a(1)+b(−1)+ca−b+c=3....(iii)
=3−a+b...(iv)
=3−a+b...(iv)
=3−a+b...(iv) 4a−2b+3−a+b=1
=3−a+b...(iv) 4a−2b+3−a+b=1
=3−a+b...(iv) 4a−2b+3−a+b=1 3a−b=−2....(v)
=3−a+b...(iv) 4a−2b+3−a+b=1 3a−b=−2....(v)
=3−a+b...(iv) 4a−2b+3−a+b=1 3a−b=−2....(v) a+b+3−a+b=2
=3−a+b...(iv) 4a−2b+3−a+b=1 3a−b=−2....(v) a+b+3−a+b=2
=3−a+b...(iv) 4a−2b+3−a+b=1 3a−b=−2....(v) a+b+3−a+b=2 2b=−1
=3−a+b...(iv) 4a−2b+3−a+b=1 3a−b=−2....(v) a+b+3−a+b=2 2b=−1
=3−a+b...(iv) 4a−2b+3−a+b=1 3a−b=−2....(v) a+b+3−a+b=2 2b=−1 b=2−1
=3−a+b...(iv) 4a−2b+3−a+b=1 3a−b=−2....(v) a+b+3−a+b=2 2b=−1 b=2−1 Replacing value of b in (v)
=3−a+b...(iv) 4a−2b+3−a+b=1 3a−b=−2....(v) a+b+3−a+b=2 2b=−1 b=2−1 Replacing value of b in (v)
=3−a+b...(iv) 4a−2b+3−a+b=1 3a−b=−2....(v) a+b+3−a+b=2 2b=−1 b=2−1 Replacing value of b in (v) 3a+21=−2
=3−a+b...(iv) 4a−2b+3−a+b=1 3a−b=−2....(v) a+b+3−a+b=2 2b=−1 b=2−1 Replacing value of b in (v) 3a+21=−2 3a=2−5
=3−a+b...(iv) 4a−2b+3−a+b=1 3a−b=−2....(v) a+b+3−a+b=2 2b=−1 b=2−1 Replacing value of b in (v) 3a+21=−2 3a=2−5 a=−6−5
Replacing value of b in (v)
Replacing value of b in (v)
Replacing value of b in (v) 3a+21=−2
Replacing value of b in (v) 3a+21=−2 3a=2−5
Replacing value of b in (v) 3a+21=−2 3a=2−5 a=−6−5
Replacing value of b in (v) 3a+21=−2 3a=2−5 a=−6−5 c=3−(6−5)+(2−1)
Replacing value of b in (v) 3a+21=−2 3a=2−5 a=−6−5 c=3−(6−5)+(2−1) c=310
Replacing value of b in (v) 3a+21=−2 3a=2−5 a=−6−5 c=3−(6−5)+(2−1) c=310 y=6−5x2−21+310.
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