Answer: B) plane FDE
C) points D, A, and E
D)
E)
<u>Step-by-step explanation:</u>
B) A plane is identified by 3 points where at least one of them is not on the same line as the other two.
C) Collinear means "same line".
D) A segment is represented by a bar over the letters of the two points.
E) Ray AD has a starting point at A and passes through D. The opposite ray will start at point A and go in the opposite direction.
Answer:
A. △P'Q'R' does not equal △P''Q''R''.
B. Reflecting across UT would change the orientation of the figure.
C. The sequence does not include a reflection that exchanges U and S.
D. Rotating about point U is not a rigid motion because it changes the orientation of the figure.
E. Translating point R' to Q' is a non-invertible transformation because it changes the location of P'.
(D) Rotating about U is not a rigid motion because it changes the orientation of the figure. [I think D is an incorrect answer choice.]
Step-by-step explanation:
Proof No.1
Jordan wants to prove △PQR≅△STU using a sequence of rigid motions. This is Jordan's proof. Translate △PQR to get △P'Q'R' with R'=U. Then rotate △P'Q'R' about point U to get △P''Q''R''. Since translation and rotation preserve distance, R''Q''=RQ=UT, and Q''=T. Reflect △P''Q''R'' across UT to get △P'''Q'''R''. Since reflection preserves distance, P'''R'''=PR=US, and P'''=S. A sequence of rigid motions maps △PQR onto △STU, so △PQR≅△STU.
Proof No.2
Jordan wants to prove △PQR≅△STU using a sequence of rigid motions. This is Jordan's proof. Translate △PQR to get △P'Q'R' with R'=U. Then rotate △P'Q'R about point U to get △P''Q''R'' so that R''Q'' and UT coincide. Since translation and rotation preserve distance, R''Q''=RQ=UT, and Q''=T. Reflect △P''Q''R'' across UT to get △P'''Q'''R''. Since reflection preserves distance, P'''R'''=PR=US, and P'''=S. A sequence of rigid motions maps △PQR onto △STU, so △PQR≅△STU.
Answer:
the solution is (-3, -2)
Step-by-step explanation:
Please, separate your equations with a (;) or a (.). Better yet, write only one equation per line:
y=2x+4
3x−6y=3
Substitute 2x + 4 for y in the second equation:
3x - 6(2x + 4) = 3, or
3x - 12x - 24 = 3
Combining like terms, we get -9x = 27, and so x = -27/9, or x = -3.
Since we know that y=2x+4, we replace x in this equation with -3 and calculate y:
y = 2(-3) + 4 = -2
Then the solution is (-3, -2).