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Karolina [17]
3 years ago
9

Write the quadratic equation whose roots are six and -2 and who’s leading coefficient is 3 use the letter X to represent the var

iable
Mathematics
1 answer:
Artist 52 [7]3 years ago
6 0

Answer:

3X^{2} - 12X - 36 = 0

Step-by-step explanation:

6 and - 2 are the only two solutions to the required quadratic equation.

So, if the variable is represented by X then (X - 6) and (X + 2) will be the only two factors of the polynomial function.

Therefore, the equation is  

(X - 6)(X + 2) = 0

⇒ X^{2} - 4X - 12 = 0

If the leading coefficient of the equation is 3 then we can write the equation as 3X^{2} - 12X - 36 = 0 (Answer)

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find the work done by a person pushing a 40 pound box up an incline that is 30 degrees above the horizontal and 100 feet long (i
tankabanditka [31]

The work done in pushing the box is 64000 lb-ft²/s²

<h3>How to calculate the work done by the person?</h3>

Since person pushing a 40 pound box up an incline that is 30 degrees above the horizontal and 100 feet long, the work done, W is

W = mgh where

  • m = mass of box = 40 lb,
  • g = acceleration due to gravity = 32 ft/s² and
  • h = height of incline = LsinФ where
  • L = length of incline = 100 ft and
  • Ф = angle of incline = 30°

So, W = mgh

W = mgLsinФ

So, substituting the values of the variables into the equation, we have

W = mgLsinФ

W = 40 lb × 32 ft/s² × 100 ftsin30°

W = 1280 lb-ft/s² × 100 ft × 0.5

W = 1280 lb-ft/s² × 50 ft

W = 64000 lb-ft²/s²

So, the work done in pushing the box is 64000 lb-ft²/s²

Learn more about work done here:

brainly.com/question/25970931

#SPJ1

3 0
1 year ago
Help with Pre Calculus
marshall27 [118]

Answer:

The first option

Step-by-step explanation:

The domain of a rational function should be all real numbers except for when the denominator is equal to 0. To find when the denominator is equal to 0 you simply need to find the zeroes of the denominator... but in this case you can do that through factoring and using the quadratic equation.

So first step is going to be to factor out the GCF, which in this case is x. This gives you the equation. x(2x^2-x-15). So one of the zeroes is when x=0. Now to find the other two zeroes you can use the quadratic equation which is x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\. So to find the other zeroes you simply plug the values in. a=2, b=-1, c=-15

x=\frac{-(-1)\pm\sqrt{(-1)^2-4(2)(-15)}}{2(2)}\\\\x=\frac{1\pm\sqrt{1-4(2)(-15)}}{4}\\\\x=\frac{1\pm\sqrt{121)}}{4}\\\\x=\frac{1\pm11}{4}\\\\x=\frac{12}{4}\\x=3\\\\x=\frac{-10}{4}\\x=-\frac{5}{2}

4 0
2 years ago
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