The given quadratic describes a parabola that opens upward. Its one absolute extreme is a minimum that is found at x = -3/2. The value of the function there is
(-3/2 +3)(-3/2) -1 = -13/4
The one relative extreme is a minimum at
(-1.5, -3.25).
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For the parabola described by ax² +bx +c, the vertex (extreme) is found where
x = -b/(2a)
Here, that is x=-3/(2·1) = -3/2.
Answer:
y=−4/3x+10
Step-by-step explanation:
Find the negative reciprocal of the slope of the original line and use the slope-intercept form y=mx+b to find the line perpendicular to 3x−4y=−2.
Answer:
Part a) <1=72°
Part b) <2=108°
Part c) <3=72°
Part d) <4=108°
Step-by-step explanation:
step 1
Find the measure of angle 1
we know that
<1+108°=180° -----> by supplementary angles
so
<1=180°-108°=72°
step 2
Find the measure angle 2
we know that
<2=108° -----> by corresponding angles
step 3
Find the measure angle 3
we know that
<3=<1-----> by corresponding angles
so
<3=72°
step 4
Find the measure angle 4
we know that
<4=108° -----> by alternate exterior angles
Answer:
-2.46, -10/4, 14/5, 2.5.
Step-by-step explanation: