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3 years ago

[ MATH, I NEED HELP PLEASE ]

Mathematics

2 answers:

miv72 [106K]3 years ago

7 0

Answer:

Line z2 : Point L

Line z1 : Point A

Step-by-step explanation:

To find the complex conjugate you flip the sign of the complex number but keep the same magnitudes.

For line z2 we can see that one of the points is $6-7i$ . To find the complex conjugate we just flip the sign. So the complex conjugate is $6+7i$ .

This means that the complex conjugate of line z2 can be represented with point: L

For line z1 we do the same thing. $1-2i$ goes to $1+2i$ . This can be represented with the point A

sashaice [31]3 years ago

4 0

The person who answered is correct!

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Solve for s: -2s < 6

Rudik [331]

Answer:

s > -3

Step-by-step explanation:

-2s < 6

Divide both sides of the inequality by -2. Remember that when you multiply or divide both sides of an inequality by a negative number, the inequality sign changes direction.

$\dfrac{-2s}{-2} > \dfrac{6}{-2}$

s > -3

Answer: s > -3

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3 years ago

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4 years ago

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4 0

3 years ago

D/d{cosec^-1(1+x²/2x)} is equal to

SIZIF [17.4K]

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg)$

Let assume that

$\rm :\longmapsto\:y = {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg)$

We know,

$\boxed{\tt{ {cosec}^{ - 1}x = {sin}^{ - 1}\bigg( \dfrac{1}{x} \bigg)}}$

So, using this, we get

$\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2x}{1 + {x}^{2} } \bigg)$

Now, we use Method of Substitution, So we substitute

$\red{\rm :\longmapsto\:x = tanz \: \rm\implies \:z = {tan}^{ - 1}x}$

So, above expression can be rewritten as

$\rm :\longmapsto\:y = sin^{ - 1} \bigg( \dfrac{2tanz}{1 + {tan}^{2} z} \bigg)$

$\rm :\longmapsto\:y = sin^{ - 1} \bigg( sin2z \bigg)$

$\rm\implies \:y = 2z$

$\bf\implies \:y = 2 {tan}^{ - 1}x$

So,

$\bf\implies \: {cosec}^{ - 1}\bigg( \dfrac{1 + {x}^{2} }{2x} \bigg) = 2 {tan}^{ - 1}x$

Thus,

$\rm :\longmapsto\:\dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg)$

$\rm \: = \: \dfrac{d}{dx}(2 {tan}^{ - 1}x)$

$\rm \: = \: 2 \: \dfrac{d}{dx}( {tan}^{ - 1}x)$

$\rm \: = \: 2 \times \dfrac{1}{1 + {x}^{2} }$

$\rm \: = \: \dfrac{2}{1 + {x}^{2} }$

Hence, 

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} {cosec}^{ - 1} \bigg( \dfrac{1 + {x}^{2} }{2x} \bigg) = \frac{2}{1 + {x}^{2} }}}}$

Hence, Option (d) is correct.

6 0

3 years ago

What is the explicit formula for -5, 10, -20, 40,...

algol [13]

You multiply by -2 every time.

3 0

3 years ago