Answer:
The answer would be c
Step-by-step explanation:
Answer:
By S.S.S. congruence property; ΔTZW ≅ ΔVZU
Step-by-step explanation:
Given:
TUVW is a rectangle.
To Prove : TZW ≅ UZV
Proof:
Since TUVW is a rectangle, and we know that opposite side of a rectangle is equal.
So,
And also TV and WU are the diagonals of the rectangle.
And the diagonals of rectangle bisects each other.
Therefore;
Now In ΔTZW and ΔVZU
TW = UV (from 1)
TZ = ZV (from 2)
WZ = ZU (from 3)
So, by S.S.S. congruence property;
ΔTZW ≅ ΔVZU
Hence proved.
Easy, it says "for ANY positive integer" so just test any positive integer
remember that n! means times all integers from 1 to that number n
lets try 1
(1+1)!/(1!)-1=
(2!)/(1!)-1=
2/1-1=
2-1=
1
if you don't believe me, try 2
(2+1)!/(2!)-2=
(3!)/(2!)-2=
(6)/(2)-2=
3-2=
1
te answer is 1, B
and number 9
easy, remember the exponential law
(x^m)(x^n)=x^(m+n)
jsut add the exponents
first gropu like bases
(r^2r^2/3)(t^1/2t^-3/2)
add bases
(r^2 and 2/3)(t^-1)=
(r^2∛(r^2))(1/t)=
Answer is 264
You just multiply at three numbers
First box is EF.
Second box is segment congruence postulate.
Third box is segment additon postulate.
Fourth box is DF. For this one the last sentence basically gives you the answer.
Just so you know for the fourth I guessed on if it's DF lined or DF unlined. I made my educated guess on the fact that the last line doesn't have a line. I hope this helps, and please tell me if I got something wrong, or my explanation wasn't sufficent enough for you.