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Artyom0805 [142]
3 years ago
14

Factor tree of short division 999​

Mathematics
1 answer:
Vlada [557]3 years ago
4 0
I think it is 1! I did it like this. 9•9• 9=1

I put period but it is the division sign
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Find the equation of a line that passes through points (1, -8) and (4, 4).
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6 0
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-4x -7 - 3x + 4 = 25
zimovet [89]

Answer:

x=-4

Step-by-step explanation:

-4x -7 - 3x + 4 = 25

-7x=25-4+7

x=-28/7

x=-4

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Show each step along the Way<br><br>Ur mom
LekaFEV [45]
2x+6 = 20
take 6 from both sides
2x = 14
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8 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
Use the discriminant to describe the roots of each equation. Then select the best description. 2 = x2 + 5x. A) Double root B) Re
fredd [130]

\bf 2=x^2+5x\implies 0=x^2+5x-2\\\\\\\qquad \qquad \qquad \textit{discriminant of a quadratic}\\\\\\0=\stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{+5}x\stackrel{\stackrel{c}{\downarrow }}{-2}~~~~~~~~\stackrel{discriminant}{b^2-4ac}=\begin{cases}0&\textit{one solution}\\positive&\textit{two solutions}\\negative&\textit{no solution}\end{cases}\\\\\\(5)^2-4(1)(-2)\implies 25+8\implies 33


so we have a 33, namely two real solutions for that quadratic.


usually that number goes into a √, if you have covered the quadratic formula, you'd see it there, namely that'd be equivalent to √(33), now 33 is a prime number, and √(33) is yields an irrational value, specifically because a prime number is indivisible other than by itself or 1.


so 33 can only afford us two real irrational roots.

3 0
3 years ago
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