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Sedaia [141]
3 years ago
6

What is the distance from the origin to point P graphed on the complex plane below

Mathematics
1 answer:
anastassius [24]3 years ago
8 0

Answer:

b) √29

Step-by-step explanation:

Given: The origin is (0, 0), the point P is (2, -5).

We need to use the distance formula.

Distance formula =

Here x2 = 2, x 1 = 0, y1 =0, y2 = -5

Distance =

= √(4 + 25)

= √29

The answer is b) √29

Hope you will understand the concepts

Thank you!!

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<span>Solve.

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</span>x^2 + 4x + 4 = 18 \\  \\ x^2+4x+4-18= 0 \\  \\ x^2+4x-14=0 \\  \\ x_1_y_2=  \dfrac{-b\pm \sqrt{b^2-4ac} }{2a} \qquad a= 1\qquad b= 4\qquad c= -14 \\  \\  \\  x_1_y_2=  \dfrac{-4\pm \sqrt{4^2-4(1)(-14)} }{2(1)} \\  \\  \\  x_1_y_2=  \dfrac{-4\pm \sqrt{16-(-56)} }{2}  \\  \\  \\  x_1_y_2=  \dfrac{-4\pm \sqrt{72} }{2}  \\  \\  \\  x_1=  \dfrac{-4+ \sqrt{72} }{2} \qquad\qquad x_2=  \dfrac{-4+ \sqrt{72} }{2}\\  \\  \\  x_1=  \dfrac{-4+ \sqrt{2^3*3^2} }{2} \qquad\qquad x_2=  \dfrac{-4- \sqrt{2^3*3^2} }{2}
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</span>x_1=  \dfrac{-4+2*3 \sqrt{2} }{2} \qquad\qquad x_2=  \dfrac{-4- 2*3\sqrt{2} }{2} \\  \\  \\ x_1=  \dfrac{-4+6 \sqrt{2} }{2} \qquad\qquad x_2=  \dfrac{-4- 6\sqrt{2} }{2} \\  \\  \\ x_1=  \dfrac{-4}{2} + \dfrac{6 \sqrt{2} }{2} \qquad\qquad x_2=   \dfrac{-4}{2}- \dfrac{ 6\sqrt{2} }{2} \\  \\  \\  x_1= -2 + 3 \sqrt{2} \qquad\qquad\quad  x_2=  -2- 3\sqrt{2}  \\  \\  \\ \boxed{x=   -2\pm 3\sqrt{2} }  \to D)<span>
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7 0
3 years ago
Read 2 more answers
I need this answered, please and thanks
Nady [450]

Answer:

P = \frac{28\sqrt{2} }{2} units

A = 24.5 unit^{2}

Step-by-step explanation:

Length of one side

sin 45^{o} = \frac{x}{7}

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7(\frac{\sqrt{2} }{2})= x

\frac{7\sqrt{2} }{2} = x

Perimeter = 4(\frac{7\sqrt{2} }{2}) = \frac{28\sqrt{2} }{2}

Area = (\frac{7\sqrt{2} }{2})(\frac{7\sqrt{2} }{2}) = \frac{98}{4} =24.5

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