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3 years ago

Help me plz f(x)=x^3+6x^2+x^1/2

Mathematics

1 answer:

Andreas93 [3]3 years ago

6 0

Answer:

Step-by-step explanation:

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3 years ago

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Nataly [62]

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2 years ago

Please help thanks !

Kitty [74]

Answer:

option D

Step-by-step explanation:

$x^{2} + y^{2} = 16\\\\x^{2} = 16- y^{2}$

equation 2:

$\frac{x^{2}}{4} - \frac{y^{2}}{25} = 1$

so we have:

$\frac{16- y^{2} }{4}-\frac{y^{2}}{25}= 1$

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3 years ago

When the sum of \, 528 \, and three times a positive number is subtracted from the square of the number, the result is \, 120. F

aleksandr82 [10.1K]

Let $x$ be the unknown number. So, three times that number means $3x$ , and the square of the number is $x^2$

We have to sum 528 and three times the number, so we have $528+3x$

Then, we have to subtract this number from $x^2$ , so we have

$x^2-(3x+528)$

The result is 120, so the equation is

$x^2 - 3x - 528 = 120 \iff x^2 - 3x - 648 = 0$

This is a quadratic equation, i.e. an equation like $ax^2+bx+c=0$ . These equation can be solved - assuming they have a solution - with the following formula

$x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

If you plug the values from your equation, you have

$x_{1,2} = \dfrac{3\pm\sqrt{9-4\cdot(-648)}}{2} = \dfrac{3\pm\sqrt{9+2592}}{2} = \dfrac{3\pm\sqrt{2601}}{2} = \dfrac{3\pm51}{2}$

So, the two solutions would be

$x = \dfrac{3+51}{2} = \dfrac{54}{2} = 27$

$x = \dfrac{3-51}{2} = \dfrac{-48}{2} = -24$

But we know that x is positive, so we only accept the solution $x = 27$

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3 years ago