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2 years ago

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The force needed to stop a car varies directly as its weight W and the square root of velocity V are inversely as the distance d

travelled. When F =36 Newtons, W=120kg, V=100km/h and d=50km. Calculate the force required for V=256km/h, W=200kg and d=800km

1 answer:

svet-max [94.6K]2 years ago

8 0

Answer:

Step-by-step explanation:

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3 years ago

Use the discriminant to predict the nature of the solutions to the equation 4x-3x²=10. Then, solve the equation.

AleksandrR [38]

Answer:

Two imaginary solutions:

x₁= $\frac{2}{3} -\frac{1}{3} i\sqrt{26}$

x₂ = $\frac{2}{3} +\frac{1}{3} i\sqrt{26}$

Step-by-step explanation:

When we are given a quadratic equation of the form ax² +bx + c = 0, the discriminant is given by the formula b² - 4ac.

The discriminant gives us information on how the solutions of the equations will be.

<u>If the discriminant is zero</u>, the equation will have only one solution and it will be real
<u>If the discriminant is greater than zero</u>, then the equation will have two solutions and they both will be real.
<u>If the discriminant is less than zero,</u> then the equation will have two imaginary solutions (in the complex numbers)

So now we will work with the equation given: 4x - 3x² = 10

First we will order the terms to make it look like a quadratic equation ax²+bx + c = 0

So:

4x - 3x² = 10

-3x² + 4x - 10 = 0 will be our equation

with this information we have that a = -3 b = 4 c = -10

And we will find the discriminant: $b^{2} -4ac = 4^{2} -4(-3)(-10) = 16-120=-104$

Therefore our discriminant is less than zero and we know<u> that our equation will have two solutions in the complex numbers. </u>

To proceed to solve the equation we will use the general formula

x₁= (-b+√b²-4ac)/2a

so x₁ = $\frac{-4+\sqrt{-104} }{2(-3)} \\\frac{-4+\sqrt{-104} }{-6}\\\frac{-4+2\sqrt{-26} }{-6} \\\frac{-4+2i\sqrt{26} }{-6} \\\frac{2}{3} -\frac{1}{3} i\sqrt{26}$

The second solution x₂ = (-b-√b²-4ac)/2a

so x₂= $\frac{-4-\sqrt{-104} }{2(-3)} \\\frac{-4-\sqrt{-104} }{-6}\\\frac{-4-2\sqrt{-26} }{-6} \\\frac{-4-2i\sqrt{26} }{-6} \\\frac{2}{3} +\frac{1}{3} i\sqrt{26}$

These are our two solutions in the imaginary numbers.

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