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3 years ago

6x^3+3n^2+8n+4 factor the following polynomial

Mathematics

1 answer:

Archy [21]3 years ago

6 0

Answer:

x1=((2√3)/3)i

x2=-0.5

x3=-((2√3)/3)i

Step-by-step explanation:

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ANEK [815]

Answer:

its the last one hun

Step-by-step explanation:

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2 years ago

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What is the slope–intercept form of the equation 4x − 2y = 9?

adelina 88 [10]

Hey there! :D

Slope intercept form: y=mx+b

4x-2y=9

Add 2y to the other side.

4x=9+2y

Subtract 9 on both sides.

4x-9=2y

Flip the equation over.

2y=4x-9 <== slope intercept form

I hope this helps!
~kaikers

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3 years ago

You're playing a game where you defend your village from an orc invasion. There are 333 characters (elf, hobbit, or human) and 5

antiseptic1488 [7]

Answer:

8/15

Step-by-step explanation:

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3 years ago

Select "Growth" or "Decay" to classify each function.

allochka39001 [22]

1. The growth rate equation has a general form of:

y = A (r)^t

The function is growth when r≥1, and it is a decay when r<1. Therefore:

y=200(0.5)^2t --> Decay

y=1/2(2.5)^t/6 --> Growth

y=(0.65)^t/4 --> Decay

2. We rewrite the given equation (1/3)^d−5 = 81

Take the log of both sides:

(d – 5) log(1/3) = log 81

d – 5 = log 81 / log(1/3)

d – 5 = - 4

Multiply both sides by negative 1:

- d + 5 = 4

So the answer is D

5 0

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) = $\frac{3}{2}$

5 0

3 years ago