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2 years ago

Which are correct representations of the inequality –3(2x – 5) < 5(2 – x)? Check all that apply.

Mathematics

2 answers:

k0ka [10]2 years ago

5 0

Answer:

C. $-6x+15$

A. First Picture

Step-by-step explanation:

We are given the inequality, $-3(2x-5)$ .

On simplifying the inequality, we get,

$-3(2x-5)$

i.e. $-6x+15$

i.e. $-x$

i.e. $x>5$

So, we get the correct representations of the inequality are,

C. $-6x+15$

A. the First Picture

dimulka [17.4K]2 years ago

4 0

–6x + 15 < 10 – 5x

pic a

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3 years ago

Cos^2x+cos^2(120°+x)+cos^2(120°-x) i need this asap. pls help me

o-na [289]

Answer:

$\frac{3}{2}$

Step-by-step explanation:

Using the addition formulae for cosine

cos(x ± y) = cosxcosy ∓ sinxsiny

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= - cos60cosx - sin60sinx

= - $\frac{1}{2}$ cosx - $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos² (120 + x)

= $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

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cos(120 - x) = cos120cosx + sin120sinx

= -cos60cosx + sin60sinx

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squaring to obtain cos²(120 - x)

= $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

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Putting it all together

cos²x + $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x + $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

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Step-by-step explanation: