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Rudik [331]
2 years ago
12

Find the areas of the trapezoids. PLZ HELP!!!!

Mathematics
1 answer:
barxatty [35]2 years ago
4 0

Answer:

<em>Type 7.5 in the box!</em>

Step-by-step explanation:

Consider the trapezoid formula ( ( base 1 + base 2 ) / 2 ) * height;

Length of Base 1 - 1 unit,\\\\Length of Base 2 - 4 units,\\\\Length of Height - 3 units ( length of x - intercept in shaded region )\\\\Area of Trapezoid = ( Average of Bases ) * Height = ( ( 1 + 4 ) / 2 ) * 3 = ( 5 / 2 ) * 3 = 7.5 square units,\\\\Solution; 7.5 square units

<em>Area = 7.5 square units</em>

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Factor x^2+3x-4 using the AC method

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igomit [66]

Answer:

Vertical A @ x=3 and x=1

Horizontal A nowhere since degree on top is higher than degree on bottom

Slant A @ y=x-1  

Step-by-step explanation:

I'm going to look for vertical first:

I'm going to factor the bottom first:  (x-3)(x-1)

So we have possible vertical asymptotes at x=3 and at x=1

To check I'm going to see if (x-3) is a factor of the top by plugging in 3 and seeing if I receive 0 (If I receive 0 then x=3 gives me a hole)

3^3-5(3)^2+4(3)-25=-31 so it isn't a factor of the top so you have a vertical asymptote at x=3

Let's check x=1

1^3-5(1)^2+4(1)-25=-25 so we have a vertical asymptote at x=1 also

There is no horizontal asymptote because degree of top is bigger than degree of bottom

There is a slant asympote because the degree of top is one more than degree of bottom (We can find this by doing long division)

                       x   -1

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

x^2-4x+3 |      x^3-5x^2+4x-25

                  - ( x^3-4x^2+3x)

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

                            -x^2 +x  -25

                       -   (-x^2+4x-3)

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

                                   -3x-22

So the slant asymptote is to x-1

8 0
3 years ago
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