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3 years ago

Solve g(-1)=x^2-5x+2

Mathematics

1 answer:

olga55 [171]3 years ago

5 0

Answer:

x = 5/2 + sqrt(17)/2 or x = 5/2 - sqrt(17)/2

Step-by-step explanation:

Solve for x:

x^2 - 5 x + 2 = 0

Subtract 2 from both sides:

x^2 - 5 x = -2

Add 25/4 to both sides:

x^2 - 5 x + 25/4 = 17/4

Write the left hand side as a square:

(x - 5/2)^2 = 17/4

Take the square root of both sides:

x - 5/2 = sqrt(17)/2 or x - 5/2 = -sqrt(17)/2

Add 5/2 to both sides:

x = 5/2 + sqrt(17)/2 or x - 5/2 = -sqrt(17)/2

Add 5/2 to both sides:

Answer: x = 5/2 + sqrt(17)/2 or x = 5/2 - sqrt(17)/2

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5. A sum of money is divided among Ethan, Jun Wei

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Answer:

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Step-by-step explanation:

Let E be the amount of money Ethan starts with, J be the amount Jun Wei starts with, and R be the amount Raj starts with. Let M be the total amount of money to start with.

To start:

$E:J:R = 5:6:9$

This means Ethan has 5/20 of M, the money to start with. This reduces to 1/4 of M.

Afterwards:

$(E-50):J:R = 3:4:6$

This means Ethan has 3/13 of the remaining money (which is M-50!) after giving up 50. So

E = (M/4), or 4E = M

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$(E - 50) = 3/13 * (M - 50)$

We know 4E = M, so we can substitute that in:

$(E - 50) = 3/13 * (4E - 50)E - 50 = (12/13)E - 150/13E - 650/13 = (12/13)E - 150/13(1/13)E = 500/13E = 500$

That's how much money Ethan had to start with. But we're asked how much he has AFTER giving $50 away, so that's $450.

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Using a trigonometric identity, it is found that the values of the cosine and the tangent of the angle are given by:

$\cos{\theta} = \pm \frac{2\sqrt{2}}{3}$

$\tan{\theta} = \pm \frac{\sqrt{2}}{4}$

<h3>What is the trigonometric identity using in this problem?</h3>

The identity that relates the sine squared and the cosine squared of the angle, as follows:

$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

In this problem, we have that the sine is given by:

$\sin{\theta} = \frac{1}{3}$

Hence, applying the identity, the cosine is given as follows:

$\cos^2{\theta} = 1 - \sin^2{\theta}$

$\cos^2{\theta} = 1 - \left(\frac{1}{3}\right)^2$

$\cos^2{\theta} = 1 - \frac{1}{9}$

$\cos^2{\theta} = \frac{8}{9}$

$\cos{\theta} = \pm \sqrt{\frac{8}{9}}$

$\cos{\theta} = \pm \frac{2\sqrt{2}}{3}$

The tangent is given by the sine divided by the cosine, hence:

$\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$

$\tan{\theta} = \frac{\frac{1}{3}}{\pm \frac{2\sqrt{2}}{3}}$

$\tan{\theta} = \pm \frac{1}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$\tan{\theta} = \pm \frac{\sqrt{2}}{4}$

More can be learned about trigonometric identities at brainly.com/question/24496175

#SPJ1

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