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3 years ago

Round to the nearest hundredth O(5x + 2) = 2(9x - 6) Type your answer...

Mathematics

1 answer:

Mumz [18]3 years ago

3 0

Answer:

Solving the expression and round to the nearest hundredth

$0(5x+2)=2(9x-6)$ we get $\mathbf{x=0.67}$

Step-by-step explanation:

We need to solve the expression and round to the nearest hundredth

$0(5x+2)=2(9x-6)$

Solving:

$0(5x+2)=2(9x-6)$

We know that multiplying anything by zero we get zero

$0=2(9x-6)$

Switching sides

$2(9x-6)=0$

Multiplying 2 with terms inside the bracket

$18x-12=0$

Adding 12 on both sides

$18x-12+12=0+12\\18x=12$

Divide both sides by 18

$\frac{18x}{18}=\frac{12}{18}\\x=0.666$

Rounding to nearest hundredth we get:

$x=0.67$

So, solving the expression and round to the nearest hundredth

$0(5x+2)=2(9x-6)$ we get $\mathbf{x=0.67}$

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Answer:

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Step-by-step explanation:

$\cos x=\sin2x\qquad|\text{use}\ \sin\theta=2\sin\theta\cos\theta\\\\\cos x=2\sin x\cos x\qquad|\text{subtract}\ \cos x\ \text{from both sides}\\\\0=2\sin x\cos x-\cos x\\\\2\sin x\cos x-\cos x=0\qquad|\text{distribute}\\\\\cos x(2\sin x-1)=0\iff\underbrace{\cos x=0}_{(1)}\ \vee\ \underbrace{2\sin x-1=0}_{(2)}$

$(1)\\\cos x=0\Rightarrow x=\dfrac{\pi}{2}+k\pi;\ k\in\mathbb{Z}$

$(2)\\2\sin x-1=0\qquad|\text{add 1 to both sides}\\\\2\sin x=1\qquad|\text{divide sides by 2}\\\\\sin x=\dfrac{1}{2}\Rightarrow x=\dfrac{\pi}{6}+2k\pi\ \vee\ x=\dfrac{5\pi}{6}+2k\pi$

$\text{From}\ (1)\ \text{and}\ (2)\ \text{we have}\\\\x=\dfrac{\pi}{2}+k\pi\ \vee\ x=\dfrac{\pi}{6}+2k\pi\ \vee\ x=\dfrac{5\pi}{6}+2k\pi\\\\\text{We have the interval}\ x\in[0;2\pi).\ \text{Therefore the solution is:}\\\\x\in\left\{\dfrac{\pi}{6};\ \dfrac{\pi}{2}\ \dfrac{3\pi}{2};\ \dfrac{5\pi}{6}\right\}$

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4 years ago

Round to the nearest hundredth O(5x + 2) = 2(9x - 6) Type your answer...​

Round to the nearest hundredth O(5x + 2) = 2(9x - 6) Type your answer...