The rule (x, y) ⇒ (x-2, y+3) will translate the way you want.
Answer:
The correct option is B.
Step-by-step explanation:
Given information: AB\parallel DCAB∥DC and BC\parallel ADBC∥AD .
Draw a diagonal AC.
In triangle BCA and DAC,
AC\cong ACAC≅AC (Reflexive Property of Equality)
\angle BAC\cong \angle DCA∠BAC≅∠DCA ( Alternate Interior Angles Theorem)
\angle BCA\cong \angle DAC∠BCA≅∠DAC ( Alternate Interior Angles Theorem)
The ASA (Angle-Side-Angle) postulate states that two triangles are congruent if two corresponding angles and the included side of are congruent.
By ASA postulate,
\triangle BCA\cong \triangle DAC△BCA≅△DAC
Therefore option B is correct
The answer for number A would be 1/3
Find the line that is normal to the parabola at the given point
remember that normal means perpendicular
perpendicular lines have slopes that multiply to -1
we can use point slope form to write the equation of the line since we are given the point (1,0)
we just need the slope
take derivitive
y'=1-2x
at x=1
y'=1-2(1)
y'=1-2
y'=-1
the slope is -1
the perpendicular of that slope is what number we can multiply to get -1
-1 times what=-1?
what=1
duh
so
point (1,0) and slope 1
y-0=1(x-1)
y=x-1 is da equation
solve for where y=x-1 and y=x-x² intersect
set equatl to each other since equal y
x-1=x-x²
x²-1=0
factor difference of 2 perfect squares
(x-1)(x+1)=0
set to zero
x-1=0
x=1
we got this point already
x+1=0
x=-1
sub back
y=-1-(-1)²
y=-1-(1)
y=-1-1
y=-2
it intersects at (-1,-2)
