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4 years ago

If G-1(x) is the inverse of G(x), what is the value of G-^1(G(x))?

Mathematics

2 answers:

vlada-n [284]4 years ago

8 0

Answer:

Step-by-step explanation:

If you multiply a function and its inverse, you will will always get x back.

f-1( f(x) ) = x

kodGreya [7K]4 years ago

6 0

Answer:

the value of G^-1(G(x)) is x

Step-by-step explanation:

We define G^-1(x) as the inverse of G(x), when we talk about functions inverse does not work like with the numbers, this means that G^-1(x) is not equal to 1/G(x).

Inverse means that if:

G(x) = y

then:

G^-1(y) = x

then if we have:

G^-1(G(x)) = G^-1(y) = x

so when we evaluate a given function in its inverse, the result is the input of the function.

And is the same in the case:

G(G^-1(x)) = x

You might be interested in

If the area of a baseball diamond (shape of a square) is 8100 ft squared, how would you find the distance from 1st base to 3rd b

Readme [11.4K]

Answer:

Exact distance = $90\sqrt{2}$ feet
Approximate distance = 127.2792 feet.

===========================================================

Explanation:

The area of the square is 8100 square feet, abbreviated ft^2.

Apply the square root to find the distance from any base to its adjacent counterpart (eg: from 1st to 2nd base). So we get sqrt(8100) = 90 ft as that side distance. Notice that 90*90 = 8100.

If you were to draw a line from 1st base to 3rd base, then you would split the square into two congruent right triangles. Each right triangle is isosceles (the two legs being 90 ft each).

Use the pythagorean theorem to find the hypotenuse.

$a^2 + b^2 = c^2\\\\c = \sqrt{a^2+b^2}\\\\c = \sqrt{90^2+90^2}\\\\c = \sqrt{2*90^2}\\\\c = \sqrt{2}*\sqrt{90^2}\\\\c = \sqrt{2}*90\\\\c = 90\sqrt{2} \ \text{ ... exact distance}\\\\c \approx 127.2792 \ \text{ ... approximate distance}\\\\$

The distance from 1st to 3rd base is roughly 127.2792 feet.

5 0

2 years ago

GIVING BRAINILEST-

Nutka1998 [239]

Answer:

To construct a dilation, we need the center of dilation, which is the point from which we make the image smaller or larger. We also need a scale factor, which is the number that determines how large or small we are making the new image. If the scale factor is greater than 1, we have an enlargement, because the new image will be larger than the original image. If the scale factor is less than 1, we have a reduction because the new image will be smaller than the original image.

Enlargement

Step-1

Let's dilate triangle PQR by a scale factor of 2 with a center of dilation at point C.

Step-2

First, we will use a straightedge to connect the center of dilation C to each vertex. We are going to extend those lines across the whole page.

Step-3

Next, we'll place the point of the compass on the center of dilation C and the pencil on vertex Q and measure that distance.

Step-4

Now, without changing the size of the compass, we will move the point of the compass to vertex Q, and make a mark on the line that is extended through Q.

Step-5

Notice that since we measured the distance from C to Q and then used that same distance to mark the line, we have now created a new point that is twice the distance from C that Q was.

We now repeat the process for each of the other vertices of the triangle.

Step-6

These marks represent the vertices of our dilated image. We often use the same letters with an apostrophe to show that it is the corresponding vertex.

Step-7

Finally, we use the straightedge to connect all of the new vertices.

Now,See how triangle P

′

is twice the size of the original triangle.

Reduction

Step-1

Given triangle PQR, now let's dilate the image with a center of dilation at C and with a scale factor of

Step-2

Using the straightedge, we will connect all of the vertices to the center of dilation C. In this case, since our scale factor is less than 1, we are reducing the size of the triangle.

Step-3

The green lines are the new lines that we have drawn.

Since we have to reduce the size of the triangle by half, we want to cut the distance in half from each vertex to the center of dilation. In order to do this, we will set our compass to a distance that is more than halfway between vertex P and point C but less than the total distance.

Step-4

Keeping the compass at that same distance, start with the point of the compass on vertex P, then draw a curve that intersects the line between P and C. Next, keeping the compass at the same distance, place the point of the compass on C and draw a curve that intersects the line between P and C and intersects the previously drawn curve.

Step-5

Now, we will take the straightedge and connect the points where the curves overlap.