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4 years ago

Given f(x)= 2x²+4x-6 and g(x)=2x-2, find each function

1 answer:

5 0

F(x)= 2x²+4x-6 and g(x)=2x-2, find each function
1. (f/g) (x) = f(x)÷g(x) = (2x²+4x-6)÷(2x-2)
First factor both top and bottom:
(2x-2)(x+3)÷(2x-2) = x+3

2. f(a + 2) = plug (a+2) in anywhere there is an x in f(x)=2x²+4x-6 -->
2(a+2)^2 +4(a+2)-6 = 2(a^2+4a+4)+4a+8-6, now distribute:
2a^2+8a+8+4a+2, combine like terms
2a^2+12a+10

3. g(a/2) = plug (a/2) in anywhere there's an x in g(x)=2x-2:
2(a/2)-2 = a-2

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Pls help this is confusing :/

Aleksandr [31]

The coordinates of triangle U'V'W' include U'(8, 3), V'(4, -8) and W'(-8, -6) and this is represented by graph A shown in the image attached below.

<h3>What is a transformation?</h3>

A transformation can be defined as the movement of a point on a cartesian coordinate from its original (initial) position to a new location.

<h3>The types of transformation.</h3>

In Geometry, there are different types of transformation and these include the following:

Dilation

Reflection

Rotation

Translation

Based on the information provided, triangle UVW would be rotated counterclockwise through an angle of 270 degree at origin to produce triangle U'V'W', we have:

$\left[\begin{array}{ccc}0&1\\-1&0\end{array}\right]$

Therefore, the image of triangle UVW would be given by this matrix:

$\left[\begin{array}{ccc}0&1\\-1&0\end{array}\right]$ $\left[\begin{array}{ccc}-3&8&6\\8&4&-8\end{array}\right]$

Image = $\left[\begin{array}{ccc}8&4&-8\\3&-8&-6\end{array}\right]$

Based on the image above, we can logically deduce that the coordinates of triangle U'V'W' include U'(8, 3), V'(4, -8) and W'(-8, -6) and this is represented by graph A shown in the image attached below.

Read more on transformations here: brainly.com/question/12518192

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3 0

2 years ago

Please help me on this!

Tanya [424]

Answer: A. 4.
Explanation: solving graphically

6 0

2 years ago

Somebody please help me!

GREYUIT [131]

Answer:

second option, IQR = 13-6 = 7 and Range = 17-6 = 11

Step-by-step explanation:

7 0

3 years ago

Find x so that the points (x,x+1), (x+2,x+3) and (x+3,2x+4) form a right-angled triangle.

azamat

Let a, b, and c be vectors each starting at the origin and terminating at the points (x, x + 1), (x + 2, x + 3), and (x + 3, 2x + 4), respectively.

Then the vectors a - b, a - c, and b - c are vectors that point in directions parallel to each of the legs formed by the triangle with these points as its vertices.

If this triangle is to contain a right angle, then exactly one of these pairs of vectors must be orthogonal. In other words, one of the following must be true:

(a - b) • (a - c) = 0

or

(a - b) • (b - c) = 0

or

(a - c) • (b - c) = 0

We have

a - b = (x, x + 1) - (x + 2, x + 3) = (-2, -2)

a - c = (x, x + 1) - (x + 3, 2x + 4) = (-3, -x - 3)

b - c = (x + 2, x + 3) - (x + 3, 2x + 4) = (-1, -x - 1)

Case 1: If (a - b) • (a - c) = 0, then

(-2, -2) • (-3, -x - 3) = (-2)×(-3) + (-2)×(-x - 3) = 2x + 12 = 0 ==> x = -6

which would make a - c = (-3, 3) and b - c = (-1, 5), and their dot product is not zero. Then the triangles vertices are at the points (-6, -5), (-4, -3), and (-3, -8).

Case 2: If (a - b) • (b - c) = 0, then

(-2, -2) • (-1, -x - 1) = (-2)×(-1) + (-2)×(-x - 1) = 2x + 4 = 0 ==> x = -2

which would make a - c = (-3, -1) and b = (-1, 1), and their dot product is also not zero. The vertices are the points (-2, -1), (0, 1), and (1, 0).

Case 3: If (a - c) • (b - c) = 0, then

(-3, -x - 3) • (-1, -x - 1) = (-3)×(-1) + (-x - 3)×(-x - 1) = x ² + 4x + 6 = 0

but the solutions to x here are non-real, so we throw out this case.

So there are two possible values of x that make a right triangle, x = -6 and x = -2.

3 0

3 years ago

5 + 4 + 2 (-3) +7+(-5)

kirza4 [7]

<h2>PLEASE MARK BRAINLIEST!</h2>

Answer:

Let's Solve!

Step-by-step explanation:

5 + 4 + 2(-3) + 7 + (-5)

\__/ \__/

9 + (-6) + 7 + (-5)

|_______|

9 + (-11) + 7

|________|

16 + (-11)

|____|

Your answer is bolded.

I hope this helps!

3 0

3 years ago

Read 2 more answers