The points (-2,0) and (0,-6) are each on the graph of a linear equation. is (2,6) also
on the graph of this linear equation? explain your reasoning.

No, it does not. See below.

Step-by-step explanation:

Lets find out the linear equation that passes trough (-2, 0) and (0, -6).

We know every linear equation has the form: y = mx + b

Where m is the slope on the curve and b the independent term.

We know that, given 2 points (x1,y1) and (x2,y2) we can find the slope m as:

m = (y2-y1)/(x2-x1)

In our case lets replace (x1,y1) and (x2,y2) by (-2, 0) and (0, -6) (notice it could be done in the inverse sense where (-2, 0) is (x2,y2) and (0, -6) is (x1,y1) ). So, our slope is:

m = [-6 - 0] / [0 - (-2)]

m = -6/2 = -3

So, we have a downward linear function with slope -3, this is:

y =  -3x + b

Now, for finding b just replace any of the 2 points given in the equation. Lets replace (-2, 0):

0 = -3(-2) + b

0 = 6 + b

Subtracting 6 in both sides:

-6 = b

So, our independent term is -6 and the function is:

y = -3x - 6

Now lets see if this linear eqution passes trough (2,6). If it does, we can replace the values on the equation. Replacing x by 2:

y = -3(2) - 6 = = -6 - 6 = - 12

So, in our equation, we x is 2 y is -12, and not 6 as in the point (2,6). So, our equation does not passes trough (2,6)

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.