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x = 2
<em>right</em><em> </em><em>option</em><em> </em><em>is</em><em> </em>(E).
Step-by-step explanation:
f(x) = x³ - 3x² + 12 in interval [-2, 4]
{taking f'(x) by doing derivative of f(x)}
f'(x) = 3x² - 6x
.•. f'(x) = 0
0 = 3x² - 6x
0 = 3x(x - 2)
0 = x - 2
x = 2
Answer:
Determination of HYP,OPP,ADJ with respect to x.
<u>Opposite</u><u> </u><u>side</u><u> </u><u>of</u><u> </u><u>right</u><u> </u><u>angle</u><u>:</u><u>Hypotenuse</u><u>:</u><u> </u><u>AC</u>
<u>Opposite</u><u> </u><u>side</u><u> </u><u>of</u><u> </u><u>given</u><u> </u><u>angle</u><u>:</u><u> </u><u>Opp</u><u>:</u><u>BC</u>
<u>remaining</u><u> </u><u>side</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>triangle</u><u>:</u><u> </u><u>Adjacent</u><u>:</u><u>AB</u><u>.</u>
Answer:
D
Step-by-step explanation:
When you reflect a diagonal over a line of symmetry, the diagonal will land perfectly on the other diagonal (and vice versa). This suggests that one diagonal is a mirror copy of the other.
Another way to put it: The vertex points of the rectangle will swap when we reflect over a line of symmetry. A diagonal is simply the opposite vertex points joined together. So this is why the diagonals swap places (because the vertices line up perfectly when you apply the reflection).