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Brums [2.3K]
3 years ago
7

What is the quotient of (3x4 – 4x2 + 8x – 1) ÷ (x – 2)?

Mathematics
2 answers:
Gekata [30.6K]3 years ago
7 0

Answer:

The quotient is  3x^3 + 6x^2 +  8x + 24  + 47/(x - 2).

Step-by-step explanation:

We can use long division to solve this.

We need to add 0x^3 to the expression.

          3x^3 + 6x^2 +  8x + 24   <--------- Quotient.

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

x - 2 )  3x4 – 0x^3 - 4x2 + 8x – 1

          3x^4-6x^3

                    6x^3 - 4x^2

                    6x^3 -12x^2

                             8x^2 + 8x

                             8x^2 -16x

                                        24x - 1

                                         24x - 48

                                                   47  <----- remainder

Genrish500 [490]3 years ago
3 0

Answer:

3{x}^{3} + 6{x}^{2} + 8x + 24 + \frac{47}{x - 2}

Step-by-step explanation:

Since the divisor is in the form of <em>x - c</em>, use what is called Synthetic Division. Remember, in this formula, -c gives you the OPPOSITE terms of what they really are, so do not forget it. Anyway, here is how it is done:

2| 3 0 −4 8 −1

↓ 6 12 16 48

________________

3 6 8 24 47 → 3{x}^{3} + 6{x}^{2} + 8x + 24 + \frac{47}{x - 2}

You start by placing the <em>c</em> in the top left corner, then list all the coefficients of your dividend [3x⁴ - 4x² + 8x - 1]. You bring down the original term closest to <em>c</em> then begin your multiplication. Now depending on what symbol your result is tells you whether the next step is to subtract or add, then you continue this process starting with multiplication all the way up until you reach the end. Now, when the last term is 0, that means you have no remainder. Finally, your quotient is one degree less than your dividend, so that 3 in your quotient can be a 3x³, the 6x² follows right behind it, green then 8x, 24, and finally, your remainder of 47, which gets set over the divisor of x - 2,<em> </em>giving you the quotient of 3{x}^{3} + 6{x}^{2} + 8x + 24 + \frac{47}{x - 2}.

I am joyous to assist you anytime.

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PLEASE HELP QUICKLY 25 POINTS
Natalija [7]

Answer:

○ \displaystyle \pi

Step-by-step explanation:

\displaystyle \boxed{y = 3sin\:(2x + \frac{\pi}{2})} \\ y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{-\frac{\pi}{4}} \hookrightarrow \frac{-\frac{\pi}{2}}{2} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\pi} \hookrightarrow \frac{2}{2}\pi \\ Amplitude \hookrightarrow 3

<em>OR</em>

\displaystyle \boxed{y = 3cos\:2x} \\ y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\pi} \hookrightarrow \frac{2}{2}\pi \\ Amplitude \hookrightarrow 3

You will need the above information to help you interpret the graph. First off, keep in mind that although this looks EXACTLY like the cosine graph, if you plan on writing your equation as a function of <em>sine</em>, then there WILL be a horisontal shift, meaning that a C-term will be involved. As you can see, the photograph on the right displays the trigonometric graph of \displaystyle y = 3sin\:2x,in which you need to replase "cosine" with "sine", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the <em>sine</em> graph [photograph on the left], accourding to the horisontal shift formula above. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY <em>REALLY</em> ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the <em>sine</em> graph [photograph on the right] is shifted \displaystyle \frac{\pi}{4}\:unitto the right, which means that in order to match the <em>cosine</em> graph [photograph on the left], we need to shift the graph BACKWARD \displaystyle \frac{\pi}{4}\:unit,which means the C-term will be negative, and by perfourming your calculations, you will arrive at \displaystyle \boxed{-\frac{\pi}{4}} = \frac{-\frac{\pi}{2}}{2}.So, the sine graph of the cosine graph, accourding to the horisontal shift, is \displaystyle y = 3sin\:(2x + \frac{\pi}{2}).Now, with all that being said, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph WILL hit \displaystyle [-1\frac{3}{4}\pi, 0],from there to \displaystyle [-\frac{3}{4}\pi, 0],they are obviously \displaystyle \pi\:unitsapart, telling you that the period of the graph is \displaystyle \pi.Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the <em>midline</em>. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at \displaystyle y = 0,in which each crest is extended <em>three units</em> beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.

I am delighted to assist you at any time.

7 0
2 years ago
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Please show an actual answer i need help
jek_recluse [69]

Answer:

1. wire is approx. 143.96 ft

2. pole is approx. 106.55 ft

Step-by-step explanation:

Mapping the information on the SOHCAHTOA picture below:

Φ = 46

wire = hypotenuse

adjacent side = 100

CAH is most suitable:

cos 46° = 100/wire => wire = 100/cos 46 ≈ 143.96

pole = opposite side + 3

TOA is most suitable:

tan 46° = (pole-3)/100 => pole = 3 + 100 * tan 46 ≈ 106.55

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Nimfa-mama [501]
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I = 400 * 6% * 3
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<em>Also, a little side note not relating to the question:</em>

<em>Remember that \pi is spelled as "pi" not "pie." Pie is a dessert, while pi is a mathematical constant.</em>

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balu736 [363]
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