Create a number pattern that follows the rule<br>"subtract 7."

triangle fdp is reduced with a scale factor of 1/2 and a center of (0,0). Find the coordinates of the new vertices.

Mkey [24]

The new points are:
P' (-3, 0)
F' (1, 2)
D' (0, -2.5)

The shape is being transformed with a dilation of a scale factor of 1/2. This means that we take the given points and multiply them by 1/2 to find the new points.

Look on the graph for the current points and multiply each one by 1/2. You will have the new points.

4 0

3 years ago

I WILL GIVE BRAINLIEST TO WHOEVER IS CORRECT

labwork [276]

The second one should by right

6 0

3 years ago

By substituting the given values for x calculate the y value for the following equation y=x+2

yawa3891 [41]

Answer:

what are the given values for x?

Step-by-step explanation:

i don't know what the given values for x are

but if you know them, here's an example

let's say x=4

y=(4)+2

y=6

6 0

3 years ago

Let us have four distinct collinear points $a,$ $b,$ $c,$ and $d$ on the cartesian plane. the point $c$ is such that $\dfrac{ab}

Deffense [45]

Start with a line segment connecting two points, A and B. $\dfrac{DA}{BA}=3$ means DA is 3 times longer than BA. Clearly, D cannot fall between A and B because that would mean DA is shorter than BA. So there are two possible locations where D can be placed on the line relative to A and B.

But with $\dfrac{DB}{BA}=2$ , or the fact that DB is 2 times longer than BA, we can rule out one of these positions; referring to the attachment, if we place D to the left of A, then DB would be 4 times longer than BA.

Finally, $\dfrac{AB}{CB}=\dfrac12$ , so that CB is 2 times longer than AB. Again we have two possible locations for point C (it cannot fall between A and B), but one of them forces C to occupy the same point as D. However, A, B, C, D are distinct, so C must fall to the left of A.

Now let $d$ be the length of AB. Then the length of CD in terms of $d$ is $4d$ . We have the coordinates of C and D, and the distance between them is $\sqrt{(4-0)^2+(0-4)^2}=4\sqrt2$ . So

$4d=4\sqrt2\implies d=\sqrt2$

The slope of the line through C and D is

$\dfrac{0-4}{4-0}=-1$

and so the equation of the line through these points is

$y-4=-(x-0)\implies x+y=4$

So the coordinates of A are $(x,y)=(x,4-x)$ . The distance between C and A is $d=\sqrt2$ , so we have

$\sqrt{(x-0)^2+(4-x-4)^2}=\sqrt{2x^2}=|x|\sqrt2=\sqrt2\implies|x|=1$

Since A falls to the right of C (in the $x,y$ plane, not just in the sketch), we know to take the positive value $x=1$ . Then the $y$ coordinate is $y=4-1=3$ .