Well, first of all, the first statement (ABC = ADC) looks like it just says
that the two halves of the little square ... each side of the diagonal ...
are congruent. That's no big deal, and it's no help in answering the
question.
The effect of the dilation is that all the DIMENSIONS of the square
are doubled ... each side of the square becomes twice as long.
Then, when you multiply (length x width) to get the area, you'd have
Area = (2 x original length) x (2 x original width)
and that's
the same as (2 x 2) x (original length x original width)
= (4) x (original area) .
Here's an easy, useful factoid to memorize:
-- Dilate a line (1 dimension) by 'x' times . . . multiply the length by x¹
-- Dilate a shape (2 dimensions) by 'x' . . . multiply area by x²
-- Dilate a solid (3 dimensions) by 'x' . . . multiply volume by x³
And that's all the dimensions we have in our world.
_______________________________
Oh, BTW . . .
-- Dilate a point (0 dimensions) by 'x' . . . multiply it by x⁰ (1)
Answer:
Origin
Step-by-step explanation:
The points on a graph that are (0,0) where the two axes intersect is called the origin.
17.7 is the answer just divide by the whole number
An interesting question!
Instead of solving each set of non-linear system of equations 6 times, I will try a simpler way, which is more adapted to multiple choice questions. However, please be warned that this method will not improve immensely math skills, but will help with reasoning, and possibly a broader understand in system of equations.
For simplicity, I will denote the sets of system of non-linear equations by S={A,B,C,D,E,F} from left to right.
1. given solutions {(-2,3),(7,-6)}
We first check each member of S, i.e. A,B,C,D,E,F for the linear conditions.
A: x+y=-2+3=1 ≠ 3 so no
B: x-y=-2-3=-5 ≠ 1 so no
C: 2x+y=-2(2)+3=-1 ≠ 1 so no
D: x+2y=-2+3(2)=4 ≠ 2 so no
E: -x+y=2+3=5 ≠ 1 so no
F: x+y=-2+3=1 ≠ 3 so YES now check the other solution 7-6=-1 OK
Now check the non-linear condition for F:
S1: (-2,3)
y-15=3-15=-12
-x^2+4x=-(-2)^2+4*(-2)=-12 good
S2:(7,-6)
y-15=-6-15=-21
-x^2+4x=-(7^2)+4(7)=-49+28=-21 also good,
So Solution set (1) matches tile F
(2) Given solutions {(-5,8),(3,0)}
We proceed in a similar way, to find that
-5+8=3 & 3+0=3 (matches A) and nothing else.
Check non-linear conditions (optionally, I transposed the terms to make comparison easier)
A. x^2+x-y=(-5)^2+(-5)-8=25-5-8=12 (looks good)
x^2+x-y=(3)^2+(3)-0=9+3=12 (looks even better)
So we determined that {(-5,8),(3,0)} is the solution for tile A.
(3) For solution set {(-2,5),(3,-5)}
Check linear conditions:
x+y=3 and -3, so does not satisfy A,& F.
x-y=-7 and 8, so does not satisfy B
2x+y=5 and 1 so satisfies C
x+2y=8 & -7 so does not satisfy D
-x+y=7 & -8 so does not satisfy E
Thus C is our only set. NOTE: if two distinct points satisfy one linear condition (say C), then both points cannot satisfy another non-equivalent linear condition, i.e. we didn't really have to check points D & E (which are not equivalent linear conditions as C) once we have found C.
Now check non-linear conditions:
x^2-3x-y=(-2)^2-3(-2)-5=5 ok
x^2-3x-y=(3)^2-3(3)-(-5)=5 ok
So solution set (3) matches tile C.
@officiallyqueenz, for your benefit, I will leave one for you as exercise. Please proceed to find the system for solution set (4). Post a your work as comments if you get stuck, or your answer for verification if you wish.
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