We compute for the side lengths using the distance formula √[(x₂-x₁)²+(y₂-y₁)²].
AB = √[(-7--5)²+(4-7)²] = √13
A'B' = √[(-9--7)²+(0-3)²] = √13
BC = √[(-5--3)²+(7-4)²] = √13
B'C' = √[(-7--5)²+(3-0)²] =√13
CD = √[(-3--5)²+(4-1)²] = √13
C'D' = √[(-5--7)²+(0--3)²] = √13
DA = √[(-5--7)²+(1-4)²] = √13
D'A' = √[(-7--9)²+(-3-0)²] = √13
The two polygons are squares with the same side lengths.
But this is not enough information to support the argument that the two figures are congruent. In order for the two to be congruent, they must satisfy all conditions:
1. They have the same number of sides.
2. All the corresponding sides have equal length.
3. All the corresponding interior angles have the same measurements.
The third condition was not proven.
1.30 :)
All you have to do is see if the number on your left is greater than 5 or not if it’s is round up 1 if not it stays the same !
Hope this helped !!!
15 dividido en 200 es <span>0.075</span>
Thee function is given as:
<span>v(t)=t^2−2tv(t)=t^2−2t
</span>(a) What is the initial velocity?
Base from the function, the initial velocity would be zero.
(b) When does the object have a velocity of zero?
It would be during time zero. Also, at a time equal to two.
