Q: Is the point with coordinates (1529) on the line y 2x-3?

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2X3 2 + 2 + 2 = 6 3 + 3 = 6 ^ thats how ^

f'(x) = 1/(2x3 + 5) rewrite f'(x) = (2X3 + 5) -1 use the chain rule d/dx (2x3 + 5) - 1 -1 * (2x3 + 5)-2 * 6x2 - 6x2(2x3 + 5) -2 ==================I would leave like this rather than rewriting this

Rearrange: 4x5 + 6x2 + 6x3 + 9 Group: 2x2 (2x3 + 3) + 3 (2x3 + 3) Simplify to get your answer: (2x2 + 3) (2x3 + 3)

6

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2X3 2 + 2 + 2 = 6 3 + 3 = 6 ^ thats how ^

The tangent to 2x3 - 3x2 - 8x + 9 at x = 2 is y = 4x - 11 The tangent to y = 2x3 - 3x2 - 8x + 9 at x = 2 has the same gradient as the curve at that point; to find the gradient, differentiate: dy/dx = 6x2 - 6x - 8 which at x = 2 is: gradient = 6 x 22 - 6 x 2 - 8 = 4 At x = 2, y = 2 x 23 - 3 x 22 - 8 x 2 + 9 = -3 The equation of a line through point (xo, yo) with gradient m is: y - yo = m(x - xo) Thus the equation of the tangent to the line at x = 2 is: y - -3 = 4(x - 2) ⇒ y = 4x - 11

(2x3)+(3x5)-(3x2)= 2x3=6 3x5=15 3x2=6 So..... 6x25-6= 6x25=150 150+6=156

your equation is this... 2x3 + 11x = 6x 2x3 + 5x = 0 x(2x2 + 5) = 0 x = 0 and (5/2)i and -(5/2)i

f'(x) = 1/(2x3 + 5) rewrite f'(x) = (2X3 + 5) -1 use the chain rule d/dx (2x3 + 5) - 1 -1 * (2x3 + 5)-2 * 6x2 - 6x2(2x3 + 5) -2 ==================I would leave like this rather than rewriting this

21

no

False

What are the factors? 2x3 - 8x2 + 6x = 2x(x - 1)(x - 3).

Rearrange: 4x5 + 6x2 + 6x3 + 9 Group: 2x2 (2x3 + 3) + 3 (2x3 + 3) Simplify to get your answer: (2x2 + 3) (2x3 + 3)

1x6 2x3

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