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uranmaximum [27]
3 years ago
13

PLEASE HELP im dying over here

Mathematics
1 answer:
Iteru [2.4K]3 years ago
7 0
\dfrac{x}{24\sqrt{2}}=\cos45^\circ\\\\\\\dfrac{x}{24\sqrt{2}}=\dfrac{\sqrt{2}}{2}\qquad\qquad\qquad\text{(cross multiply)}\\\\\\2x=24\sqrt{2}\cdot\sqrt{2}\\\\2x=24\cdot2\quad|:2\\\\\boxed{x=24}
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\large\boxed{a(x+4)(x-5)=0,\ a\in\mathbb{R}-\{0\}}\\\\\text{for}\ a=1\to\boxed{x^2-x-20=0}

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\text{If}\ x_1\ \text{and}\ x_2\ \text{are the roots of the quadratic equation}\ ax^2+bx+c=0,\\\text{then}\ ax^2+bx+c=a(x-x_1)(x-x_2).\\\\\text{If}\ x_1\ \text{and}\ x_2\ \text{, then they are the roots of the quadratic equation.}\\\\\text{We have}\ x_1=-4\ \text{and}\ x_2=5.\ \text{Therefore we have the equation:}\\\\a(x-(-4))(x-5)=0\\\\a(x+4)(x-5)=0\qquad\text{for any value of}\ a\ \text{except 0}.

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