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1 year ago

Two functions are defined as shown. f(x) =-1/2 x – 2 g(x) = –1 Which graph shows the input value for which f(x) = g(x)?

Mathematics

2 answers:

Alchen [17]1 year ago

5 0

Answer:

The second graph

Step-by-step explanation:

Just did it!!

Natalka [10]1 year ago

4 0

Answer:

See attachment.

Step-by-step explanation:

The given functions are:

$f(x) = - \frac{1}{2}x - 2$

and g(x)=-1

To find the x-value of the point of intersection of the two functions, we equate the two functions and solve for x.

$- \frac{1}{2}x - 2 = -1$

$x + 4 = 2$

$x = 2 - 4 = - 2$

The graph that shows the input value for which f(x)=g(x) is the graph which shows the point of intersection of f(x) and g(x) to be at x=-2.

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A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si

Grace [21]

Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

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$Width = y + 4$

So:

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$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

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$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

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what is the simplified form of the following expression? Assume x is greater than or equal to 0 and y is greater than or equal t

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the given expression is :

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