Answer:
<em><u>Using the mid-term break formula for all of them.</u></em>
a) x²+7x+12
= x²+3x+4x+12
= x(x+3)+4(x+3)
= (x+4)(x+3)
b) x²+6x+8
= x²+2x+4x+8
= x(x+2)+4(x+2)
= (x+4)(x+2)
c) x²+5x+6
= x²+2x+3x+6
= x(x+2)+3(x+2)
= (x=3)(x+2)
d) x²+8x+7
= x²+7x+x+7
= x(x+7)+1(x+7)
= (x+1)(x+7)
I'm not sure I understand the question, sorry!
Let r(cos O + i sin O) be a cube root of 125(cos 288 + i sin 288)
then
r^3(cos O + i sin O)^3 = 125(cos 288 + i sin 28)
so r^3 = 125 and cos 3O + i sin 3O = cos 288 + i sin 288
so r = 5 and 3O = 288 + 360p and O = 96 + 120p
so one cube root is 5 (cos 96 + i sin 96)
Im a little rusty at this stuff Its been a long time.
Im not sure of the other 2 roots
sorry cant help you any more
Answer:
<em>a) </em>
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<em>b) The coordinates of P are</em>
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Step-by-step explanation:
<u>Translation</u>
The dashed line shows the graph of the function
This function has a maximum value of 1, a minimum value of -1, and a center value of 0.
a)
Graph G shows the same function but translated by 2 units up, thus the equation of G is:
b) The coordinates of P correspond to the value of
The value of G is
Since
The coordinates of P are
