Webb22 mars 2024 · Question 1 The abscissa of the point on the curve 3y = 6x – 5x3, the normal at which passes through origin is: 1 (B) 1/3 (C) 2 (D) 1/2 Given curve 3𝑦=6𝑥−5𝑥^3 Finding 𝒅𝒚/𝒅𝒙 𝒅𝒚/𝒅𝒙=𝟐−𝟓𝒙^𝟐 Now, Slope of normal × Slope of tangent = −1 Slope of normal (m) = (−𝟏)/(𝑺𝒍𝒐𝒑𝒆 𝒐𝒇 WebbBasic Math Examples Popular Problems Basic Math Simplify -4g+2g (3g^2+8g-6) −4g + 2g (3g2 + 8g − 6) - 4 g + 2 g ( 3 g 2 + 8 g - 6) Simplify each term. Tap for more steps... −4g+ 6g3 +16g2 −12g - 4 g + 6 g 3 + 16 g 2 - 12 g Subtract 12g 12 g from −4g - 4 g. 6g3 + 16g2 −16g 6 g 3 + 16 g 2 - 16 g
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Simplify 6h2-3g3+7+3f2+9g3-5h2-2f2-9 - Gauthmath
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