When reflecting across the Y axis, the Y values remain the same.
Now if you were reflecting across Y = 0, the x values would just be inverse ( opposite signs).
So this triangle if reflected across Y = 0 the new vertices would be (4,4) (2,3) and (5,2)
Now since the reflection line is y = -1, which is a one unit shift to the left of y = 0, subtract 1 unit from each X value.
The locations are now: A'(3,4), B'(1,3) and C'(4,2)
F(x) = -3(x + 2)(x - 5)^3 > 0
(x + 2)(x - 5) > 0
x + 2 > 0 or x - 5 < 0
x > -2 or x < 5
-2 < x < 5
Therefore, the stated interval is false.
Answer:
the answer is 7 i believe
Answer:
(Véase la explicación para mayores detalles/See the explanation for further details)
Step-by-step explanation:
a) Tres al cubo (Three to cube):



b) Seis al cubo (Six to cube):



c) Cinco al cuadrado (Five squared):



d) Ocho al cuadrado (Eight squared):



e) Dos a la quinta (Two to five):



f) Diez a la cuarta (Ten to four):


