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andreyandreev [35.5K]
3 years ago
8

F(x)=3x+2 and g(x)=-2x-4, find h(x) = f(x)-g(x)

Mathematics
1 answer:
elena-14-01-66 [18.8K]3 years ago
6 0

Answer:

5x-2

Step-by-step explanation:

Since f(x)-g(x) you take 3x+2 and subtract -2x-4.

3x+2+2x-4

5x-2

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The range is 40 and the midrange is 20.

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Find the remainder when f(x) = 5x3 + 7x + 5 is divided by x + 2.
serg [7]

Answer:

Option a) is correct  that is -49.

Step-by-step explanation:

We have been given  the  two functions in which one of them is dividend and the other one is divisor

f(x)=5x^3+7x+5  when divided by x+2

You can see the long division to find the remainder in the attachment.  

First step we do is to choose the quotient so as to cancel the first term

And proceed till you get the degree of divisor less than dividend.

Hence, the remainder we will get is -49

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Use the method of Lagrange multipliers to find the dimensions of the rectangle of greatest area that can be inscribed in the ell
Tanzania [10]

Answer:

Length (parallel to the x-axis): 2 \sqrt{2};

Height (parallel to the y-axis): 4\sqrt{2}.

Step-by-step explanation:

Let the top-right vertice of this rectangle (x,y). x, y >0. The opposite vertice will be at (-x, -y). The length the rectangle will be 2x while its height will be 2y.

Function that needs to be maximized: f(x, y) = (2x)(2y) = 4xy.

The rectangle is inscribed in the ellipse. As a result, all its vertices shall be on the ellipse. In other words, they should satisfy the equation for the ellipse. Hence that equation will be the equation for the constraint on x and y.

For Lagrange's Multipliers to work, the constraint shall be in the form: g(x, y) =k. In this case

\displaystyle g(x, y) = \frac{x^{2}}{4} + \frac{y^{2}}{16}.

Start by finding the first derivatives of f(x, y) and g(x, y)with respect to x and y, respectively:

  • f_x = y,
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This method asks for a non-zero constant, \lambda, to satisfy the equations:

f_x = \lambda g_x, and

f_y = \lambda g_y.

(Note that this method still applies even if there are more than two variables.)

That's two equations for three variables. Don't panic. The constraint itself acts as the third equation of this system:

g(x, y) = k.

\displaystyle \left\{ \begin{aligned} &y = \frac{\lambda x}{2} && (a)\\ &x = \frac{\lambda y}{8} && (b)\\ & \frac{x^{2}}{4} + \frac{y^{2}}{16} = 1 && (c)\end{aligned}\right..

Replace the y in equation (b) with the right-hand side of equation (b).

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If x = 0, the area of the rectangle will equal to zero. That's likely not a solution.

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Step-by-step explanation:

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