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mars1129 [50]
3 years ago
13

Expand And Simplify (5x+2)(2x-3)

Mathematics
1 answer:
Anna007 [38]3 years ago
5 0
(5x+2)(2x-3)=10x^-15x+4x-6 = 10x^-11x-6

 4x cu + 28x cu = x cu(4+28) =  32x cu
(^ means square)

4x cu + 24x^ = 4x^( x+6) 



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Simplify the expression 2j+4j+j+7
Alina [70]

Answer:

7j+7

Step-by-step explanation:

2j+4j+j+7

combine like terms

7j+7

Sorry I don't really know how to explain it, but you just have to combine terms with the same unit

3 0
3 years ago
Identify the value of y<br><br><br> (3y+5)=-4-8y+9+77
noname [10]

Answer:

Y=7

Step-by-step explanation:

(3y+5)= -4 -8y +9 + 77

     -5    -5            

    3y= -9 - 8y + 9 + 77

           +9          -9

    3y=  -8y + 77

    +8y    +8y

      11y = 77

      11y/11    77/11

        y = 7

1. subtract 5 from both sides

2. add -9 and +9 and you get 0

3. add 8y to both sides

4. divide both sides by 11

5. you get y=7

3 0
3 years ago
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
Read 2 more answers
In △ABC point D is the midpoint of AB , point E is the midpoint of BC , and point F is the midpoint of BE . Find area of △ABC, i
xz_007 [3.2K]

I did this   manually on  some notepaper but my scanner is not working.

The answer is 72 cm^2

3 0
3 years ago
I need to know number 37 please help I keep getting it wrong
Thepotemich [5.8K]
I just worked it out and I got 6 3/28
Im sorry if that's wrong
5 0
3 years ago
Read 2 more answers
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