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3 years ago

7

Help please, its due soon !

a1" title=" {x}^{2} + 6x + 9" alt=" {x}^{2} + 6x + 9" align="absmiddle" class="latex-formula">

2 answers:

NemiM [27]3 years ago

4 0

Answer:

(x+3) (x+3)

Step-by-step explanation:

X2+6x+9

X2+3x+3x+9

x(x+3)+3(x+3)

(x+3) (x+3)

Hey mate hope it's help you.......

liraira [26]3 years ago

3 0

Answer: x = -3

Step-by-step explanation:

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3 years ago

Cos^2x+cos^2(120°+x)+cos^2(120°-x)<br>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

---------------------------------------------------------------

cos(120 + x) = cos120cosx - sin120sinx

= - 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

--------------------------------------------------------------------

cos(120 - x) = cos120cosx + sin120sinx

= -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

--------------------------------------------------------------------------

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

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

= $\frac{3}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ (cos²x + sin²x) = $\frac{3}{2}$

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3 years ago

Tom has three times as many action figures as Jerry. If Tom has 27 action figures, how many does Jerry have?

weeeeeb [17]

Answer:

Step-by-step explanation:

3j = 27

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Jerry has 9 figures.

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