1)We have to make a line perpendicular to y=2x+3, passing through the point (0,0).
y=2x+3
m=solpe=2
The slope of the line perpedicular to y=2x+3 is m´
m´=-1/m
m´=-1/2
Point slope form:
y-y₀=m(x-x₀)
(0,0)
m=-1/2
y-0=-1/2(x-0)
y=-x/2
Therefore, the line perpendicular to y=2x+3 is y=-x/2.
2)The point on the line y=2x+3 tha is closest to the origin is the point of intersection of the two lines.
y=2x+3
y=-x/2
We can solve this system of equations by equalization method.
2x+3=-x/2
least common multiple=2
4x+6=-x
4x+x=-6
5x=-6
x=-6/5 (=-1.2)
y=-x/2
y=-(-6/5)/2=6/10=3/5 (=0,6)
Therefore: the point of the line y=2x+3 that is closest to the origin is:
(-1.2 , 0.6)


Critical points occur when

, which happens for

and

.
Check the sign of the second derivative at each critical point to determine the function's concavity at that point. If it's concave (

), then a maximum occurs; if it's convex (

), then a minimum occurs.
You have

and so


This means a maximum of

and a minimum of

.
When the coefficient of eˣ, A, is:
-1 < A < 1
The graph shrinks
Answer:
D
Step-by-step explanation:
The diagonals of a rhombus bisect the angles, thus
∠ ABC = ∠ CBD = 5x - 18 and
∠ BAC = ∠ DAC = x
The diagonals are perpendicular bisectors of each other, thus
the angles round the intersection of the diagonals = 90°
Consider the triangle on the left of the rhombus, then
5x - 18 + x + 90 = 180 ( sum of angles in a triangle )
6x + 72 = 180 ( subtract 72 from both sides )
6x = 108 ( divide both sides by 6 )
x = 18 → D