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3 years ago

Can somebody help me

Mathematics

1 answer:

Sergeeva-Olga [200]3 years ago

6 0

$cis(x)=\cos x+i\sin x\\\\cis15^o=\cos15^o+i\sin15^o\\\\\cos15^o=\dfrac{\sqrt6+\sqrt2}{4};\ \sin15^o=\dfrac{\sqrt6-\sqrt2}{4}\\\\\text{therefore}\\\\7cis15^o=7\left(\cos15^o+i\sin15^o\right)=7\cos15^o+7i\sin15^o\\\\=7\left(\dfrac{\sqrt6+\sqrt2}{4}\right)+7\left(\dfrac{\sqrt6-\sqrt2}{4}\right)i$

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Anyone help with this question please..

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Answer:

x>3

Step-by-step explanation:

then reason is x is the black bold line and x the line is larger than 3 so the equation is x>3

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allochka39001 [22]

Answer:

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Step-by-step explanation:

Before I answer these here are some good notes

If the inequality sign is < or > then the circle is going to be open

If the inequality sign is $\leq or\geq$ then the circle is going to be closed

If the inequality sign is facing the variable (ex: x>3) then the line is going to be going to the right

If the inequality sign is facing the constant (ex: x<3) then the line is going to be going to the left

p>-3 open circle going to the right

$m\leq -4$ closed circle going to the left

n<5 open circle going to the left

$h\geq 2$ closed circle going to the right

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Assuming the graph is y=-3(√2x)-4 and y=-3√(x-4) the transformation would be:

The graph is compressed horizontally by a factor of 2
x=1/2x'
y=-3(√2x)-4
y=-3(√x')-4 

moved left 4
x=x'-4
y=-3(√x)-4
y=-3(√x'-4)-4

moved down 4
y=y'-4
y=-3(√x-4)-4
y'-4=-3(√x'-4)-4
y'=-3(√x'-4)-4 +4
y'=-3(√x'-4)

Answer: C. The graph is compressed horizontally by a factor of 2, moved left 4, and moved down 4.

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