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svlad2 [7]
3 years ago
15

Find the product of (3x − 4)(2x2 + 2x − 1).

Mathematics
2 answers:
alexira [117]3 years ago
6 0

Solve:

(3x    −   4)(2x^2     +      2x    −   1)


Simplify:

(3x     −      4)(2x^2      +       2x     −    1)

=      (3x     +    −4)(2x^2     +        2x        +        −1)

=   (3x)(2x^2)    +   (3x)(2x)   +   (3x)(−1)    +    (−4)(2x2)   +    (−4)(2x)     +     (−4)(−1)

=      6x^3    +   6x^2    −   3x    −  8x^2   −   8x   +   4

Answer   =      6x^3    −   2x^2     −   11x   +   4






Hope that helps!!!!                                         : )


STALIN [3.7K]3 years ago
5 0

the answer to your question is -12





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Anika [276]

Answer:

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Step-by-step explanation:

The first train left Miami at 3:00 PM.

The second train left Miami at 8:00 PM.

Let the speed of the first train be x.

The second train travels 145 mph faster than the first, that is 145 + x

The second train overtakes the first at 10:00 PM

This means that at 10:00 PM, they were both at the same location.

That is 7 hours after the first train left and 2 hours after the second train left.

Speed is given as:

s = d / t

where d = distance traveled and t = time taken

For the first train, after 7 hours traveling at speed x:

x = d / 7

=> d = 7x _______(1)

For the second train, after 2 hours, traveling at speed 145 + x:

145 + x= d / 2

=> d = 290 + 2x ____(2)

Equating (1) and (2):

7x = 290 + 2x

=> 7x - 2x = 290

5x = 290

=> x = 290/5 = 58 mph

Therefore, the first train travels at 58 mph and the second train travels at 203 mph (58 + 145)

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3 years ago
Apply the distributive property to expand the expression.<br> 9(2+5m)
zheka24 [161]
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bija089 [108]

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nalin [4]

The equation of the tangent line is y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2}  }{a^{2} y_{0} } + y_{0}

To find the tangent to the hyperbola \frac{x^{2} }{a^{2} } - \frac{y^{2} }{b^{2} } at (x₀, y₀), we differentiate the equation implicitly to find the equation of the tangent at (x₀, y₀).

So, \frac{d}{dx} (\frac{x^{2} }{a^{2} } - \frac{y^{2} }{b^{2} }) = \frac{d0}{dx}\\\frac{d}{dx} \frac{x^{2} }{a^{2} } - \frac{d}{dx}\frac{y^{2} }{b^{2} }= 0\\\frac{2x }{a^{2} } - \frac{dy}{dx}\frac{2y }{b^{2} } = 0\\\frac{2x }{a^{2} } = \frac{dy}{dx}\frac{2y }{b^{2} } \\\frac{dy}{dx} = \frac{b^{2}x }{a^{2}y }

So, at (x₀, y₀)

\frac{dy}{dx} = \frac{b^{2} x_{0} }{a^{2} y_{0} }

So, the equation of the tangent line is gotten from the standard equation of a line in point-slope form

So, \frac{y - y_{0} }{x - x_{0} } = \frac{b^{2} x_{0} }{a^{2} y_{0} }  \\y - y_{0} = \frac{b^{2} x_{0} }{a^{2} y_{0} }(x - x_{0}) \\y = \frac{b^{2} x_{0} }{a^{2} y_{0} }(x - x_{0}) + y_{0} \\y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2}  }{a^{2} y_{0} } + y_{0}

So, the equation of the tangent line is y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2}  }{a^{2} y_{0} } + y_{0}

Learn more about equation of tangent line here:

brainly.com/question/12561220

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