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3 years ago

12

CAN SOMEONE PLZ HELP

1 answer:

FinnZ [79.3K]3 years ago

4 0

Answer:

File will help, im sure

Step-by-step explanation:

<em><u>Plz mark as brainliest if correct! Have a nice day!!! </u></em>

<em><u>-Lil G</u></em>

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Which expression is equivalent to (2x2)(3x3)(4x)2?

Lera25 [3.4K]

Given expression: $(2x^2)(3x^3)(4x)^2.$

In order to simplify it to get an equivalent expression, we need to multiply all terms.

Let us simplify $(4x)^2 = 4^2x^2 = 16x^2$

Therefore, $(2x^2)(3x^3)(4x)^2=(2x^2)(3x^3)(16x^2).$

$(2x^2)(3x^3)(16x^2)=2 \times 3 \times 16 \times (x^2 \times x^3 \times x^2).$

= $96x^{2+3+2}$

$96x^{7}$ .

<h3>Therefore, $(2x^2)(3x^3)(4x)^2$ is equivalent to $96x^{7}.$ </h3>

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3 years ago

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AnnZ [28]

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3 years ago

I don’t know how to solve this

Lady_Fox [76]

Answer:

$\theta =2\pi k,\ \ k\in Z\ \\\text{or}\ \\\theta=-\dfrac{2\pi}{3}+2\pi k,\ \ k\in Z$

Step-by-step explanation:

Given:

$\cos \theta-\sqrt{3}\sin \theta=1$

Divide this equation by 2:

$\dfrac{1}{2}\cos \theta-\dfrac{\sqrt{3}}{2}\sin \theta=\dfrac{1}{2}$

Note that

$\cos \dfrac{\pi }{3}=\dfrac{1}{2}\\ \\\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$

So, the previous equation is

$\cos \dfrac{\pi}{3}\cdot \cos \theta-\sin \dfrac{\pi}{3}\cdot \sin \theta=\dfrac{1}{2}$

Remind that

$\cos x\cos y-\sin x\sin y=\cos (x+y),$

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$\cos \left(\dfrac{\pi}{3}+\theta\right)=\dfrac{1}{2}$

The solution of this equation is

$\dfrac{\pi}{3}+\theta=\pm \arccos \dfrac{1}{2}+2\pi k,\ \ k\in Z\\ \\\dfrac{\pi}{3}+\theta=\pm \dfrac{\pi}{3}+2\pi k,\ \ k\in Z\\ \\\theta =2\pi k,\ \ k\in Z\ \text{or}\ \theta=-\dfrac{2\pi}{3}+2\pi k,\ \ k\in Z$

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3 years ago

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Alex17521 [72]

Right us acute like leebron jameson

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3 years ago

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