Answer:
disributive i think
Step-by-step explanation:
It would be 0.38 the 8 is repeating
We can use elimination for these set of systems.
First, we need to set up our variables.
Belts=b
Hats=h
Now, the situation is 6 belts and 8 hats for $140. The situation after is 9 belts and 6 hats for $132.
Let’s set up our system of equations.
6b+8h=140
9b+6h=132
We need to eliminate a variable. Since b has coefficients of 6 and 9, we can easily eliminate b by multiplying the top equation by 3 and the bottom by -2.
18b+24h=420
-18b-12h=-264
Now let’s add.
12h=156
Let’s divide to get h by itself.
156/12=13=h
So a hat costs $13. We need to put in 13 for one of the equations so we can find the cost of a belt.
9b+6(13)=132
9b+78=132
We need b by itself.
9b=54
54/9=6
Belts are $6
We can also use the first equation to check our answers.
6(6)+8(13)
36+104
140.
So, the price of a belt is $6 while the price of a hat is $13.
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.
