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user100 [1]
3 years ago
6

Functions f(x) and g(x) are shown below:

Mathematics
2 answers:
laila [671]3 years ago
7 0
 f` ( x ) = 6 x + 12
       6 x + 12 = 0       6 x = - 12       x = - 2       f ( - 2 ) 0 12 - 24 + 16 = 4       f ( x ) min = 4       g` ( x ) = 4 cos (  2 x - π )       4 cos ( 2 x - π ) = 0       cos ( 2 x - π ) = 0       2 x - π = 3π / 2       2 x = 5π /2       x = 5π/4       g ( 5π/4 ) = 2 sin ( 5π/2 - π ) + 4 = 2 ( sin 3π/2 ) + 4 = -2 + 4 = 2       g ( x ) min = 2 
mestny [16]3 years ago
7 0

Answer:

g(x) has the smallest minimum y-value

Step-by-step explanation:

f(x) is the equation of a parabola

The general form of a parabola is ax² + bx + c, if a is positive, the parabola has a minimum. The minimum is at the vertex, the x-coordinate of the vertex is calculated as follows: -b/2a.  For this case, x =-12/(2*3) = -2, which corresponds to the following function value: 3(-2)² + 12(-2) + 16 = 4

  • minimum value of f(x) = 4

g(x) is the sine function

The sine function is a periodic function which oscillates between -1 and 1

  • minimum value of g(x) = -1
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As you can see both the expressions, x+2 and -4x+7 are equal to "y". That means they are equal to each other too.

So let's equal them and solve it.

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Now we have x's value we will plug it in the first equation to find "y".

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Chris made $184 for 8 hours of work. At the same rate, how much would he make for 13 hours of work?
pashok25 [27]

Answer:

$299

Step-by-step explanation:

Make a proportion

He makes $184 for 8 hours of work, and $x for 13 hours of work

184/8=x/13

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13(184/8)=(x/13)13

The 13s on the right will cancel, leaving us with

13(23)=x

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If you don't know the first derivative of \cot, but you do for \sin and \cos, you can derive the former via the quotient rule:

\cot x=\dfrac{\cos x}{\sin x}

\implies(\cot x)'=\dfrac{\sin x(-\sin x)-\cos x(\cos x)}{\sin^2x}=-\dfrac1{\sin^2x}=-\csc^2x

or if you know the derivative of \tan:

\cot x=\dfrac1{\tan x}

\implies(\cot x)'=-(\tan x)^{-2}\sec^2x=-\dfrac{\sec^2x}{\tan^2x}=-\dfrac{\frac1{\cos^2x}}{\frac{\sin^2x}{\cos^2x}}=-\dfrac1{\sin^2x}=-\csc^2x

As for the second derivative, you can use the power/chain rules:

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or if you don't know the derivative of \csc,

\csc x=\dfrac1{\sin x}

\implies(-\csc^2x)'=\left(-(\sin x)^{-2}\right)'=2(\sin x)^{-3}(\sin x)'=\dfrac{2\cos x}{\sin^3x}

which is the same as the previous result since

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