2 years ago

PLEASE HELP ME SO I CAN MOVE ON FROM THIS TOPIC

Mathematics

1 answer:

ExtremeBDS [4]2 years ago

4 0

Answer:

3rd option

Step-by-step explanation:

( $\frac{f}{g}$ )(x)

= $\frac{f(x)}{g(x)}$

= $\frac{2x^2-5x-3}{2x^2+5x+2}$ ← factorise numerator and denominator

= $\frac{(x-3)(2x+1)}{(x+2)(2x+1)}$ ← cancel (2x + 1) on numerator/ denominator

= $\frac{x-3}{x+2}$

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Describe the nature of the roots for this equation.<br> 3x^2+x-5= 0

m_a_m_a [10]

Answer:

<h2>This equation has no natural roots.</h2>

Step-by-step explanation:

$3x^2+x-5=0\\\\\text{Use the quadratic formula for}\ ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{For the given equation:}\\\\a=3,\ b=1,\ c=-5\\\\b^2-4ac=1^2-4(3)(-5)=1+60=61\\\\x=\dfrac{-1\pm\sqrt{61}}{(2)(3)}=\dfrac{-1\pm\sqrt{61}}{6}\notin\mathbb{N}$