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QUESTION 2 2.1 Solve for x: 2.1.1 (x-2)(2x+9)=0 2.1.2 4(3^3x+1)=36 2.1.3 sqrt (2x)=p 2.1.4 -2leqslant 3x+1lt 7 (represent the solution in a number line)

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QUESTION 2
2.1 Solve for x:
2.1.1 (x-2)(2x+9)=0
2.1.2 4(3^3x+1)=36
2.1.3 sqrt (2x)=p
2.1.4 -2leqslant 3x+1lt 7 (represent the solution in a number line)

QUESTION 2 2.1 Solve for x: 2.1.1 (x-2)(2x+9)=0 2.1.2 4(3^3x+1)=36 2.1.3 sqrt (2x)=p 2.1.4 -2leqslant 3x+1lt 7 (represent the solution in a number line)

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FábioAvançado · Tutor por 1 anos

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2.1.1 To solve the equation $(x-2)(2x+9)=0$, we can set each factor equal to zero and solve for x:<br /><br />$(x-2)=0$ or $(2x+9)=0$<br /><br />Solving these equations, we get:<br /><br />$x=2$ or $x=-\frac{9}{2}$<br /><br />So the solutions are $x=2$ and $x=-\frac{9}{2}$.<br /><br />2.1.2 To solve the equation $4(3^{3x+1})=36$, we can start by dividing both sides by 4:<br /><br />$3^{3x+1}=9$<br /><br />Taking the logarithm of both sides, we get:<br /><br />$\log(3^{3x+1})=\log(9)$<br /><br />Using the property of logarithms, we can bring the exponent in front:<br /><br />$(3x+1)\log(3)=\log(9)$<br /><br />Dividing both sides by $\log(3)$, we get:<br /><br />$3x+1=\frac{\log(9)}{\log(3)}$<br /><br />Simplifying the right side, we get:<br /><br />$3x+1=2$<br /><br />Subtracting 1 from both sides, we get:<br /><br />$3x=1$<br /><br />Dividing both sides by 3, we get:<br /><br />$x=\frac{1}{3}$<br /><br />So the solution is $x=\frac{1}{3}$.<br /><br />2.1.3 To solve the equation $\sqrt{2x}=p$, we can square both sides to eliminate the square root:<br /><br />$(\sqrt{2x})^2=p^2$<br /><br />Simplifying, we get:<br /><br />$2x=p^2$<br /><br />Dividing both sides by 2, we get:<br /><br />$x=\frac{p^2}{2}$<br /><br />So the solution is $x=\frac{p^2}{2}$.<br /><br />2.1.4 To solve the inequality $-2\leqslant 3x+1\lt 7$, we can start by subtracting 1 from all parts of the inequality:<br /><br />$-3\leqslant 3x\lt 6$<br /><br />Dividing all parts of the inequality by 3, we get:<br /><br />$-1\leqslant x\lt 2$<br /><br />So the solution is $-1\leqslant x\lt 2$.<br /><br />To represent the solutions on a number line, we can mark the points $x=2$, $x=-\frac{9}{2}$, $x=\frac{1}{3}$, $x=\frac{p^2}{2}$, and the interval $-1\leqslant x\lt 2$ on the number line.
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