Answer:
A [and B]
Step-by-step explanation:
If x < 5.3 then anything higher than 5.3 makes the statement false.
16 is too high and therefore is the most obvious fallacy.
5.3 would actually make x=5.3 and therefore x wouldn't be less than 5.3 making it also false but this really depends on what your teacher is looking for.
0 is true.
-8.95 is true.
If there is any clarification in the problem you can provide I can give a more definite answer but now it is really a coinflip!
(1) Answer : 
Step 1: 
Step 2: 
In completing the square method we take coefficient of x and divide by 2 and the square it . Then add it on both sides
The coefficient of x is -6. 
= (-3)^2 = 9
The coefficient of y is -4. 
= (-2)^2 = 4
Step : 
(2) 
To find center and radius we write the equation in the form of
using completing the square form
Where (h,k) is the center and 'r' is the radius
In completing the square method we take coefficient of x and divide by 2 and the square it . Then add it on both sides
Here h= -3 and k=3 and 
so r= 4
Center is (-3,3) and radius = 4
(c) Step 1: 
Step 2: 
Step 3: 
Step 4: 
We factor out each quadratic
(x^2 + 8x + 16) = (x+4)(x+4) = 
((y^2 - 6y + 9)) = (x-3)(x-3) = 
Step 5 :
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:
b hope it helps
Step-by-step explanation:
The vertex for the first one is: (8,-1)
the answer for the second one is: n = -8 + with a - at the bottom square root of 151
the answer for the third question is: n = (-3,-5)
hope i helped please thank, rate, and give me best anyswer if you're on a desktop thank you :)
