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3 years ago

What is the standard form of (2-4i)(3+5i)/(3+i)?

Mathematics

2 answers:

Eva8 [605]3 years ago

8 0

Answer:

AKA Letter Choice E on Plato!!

german3 years ago

5 0

$\dfrac{(2-4i)(3+5i)}{3+i}=\\
\dfrac{6+10i-12i+20}{3+i}=\\
\dfrac{26-2i}{3+i}=\\
\dfrac{(26-2i)(3-i)}{9+1}=\\
\dfrac{78-26i-6i-2}{10}=\\
\dfrac{76-32i}{10}=\\
\dfrac{38}{5}-\dfrac{16}{5}i$

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Hi I also need help with this please and thank you

OverLord2011 [107]

Answer:

Step-by-step explanation:

y=mx+b where m=slope and b=y intercept

m=(y2-y1)/(x2-x1)

m=(-2-4)/(2+1)

m=-2 so far we have the slope

y=-2x+b, using point (2,-2) we can solve for b, the y intercept

-2=-2(2)+b

-2=-4+b

2=b so we have our line

y=-2x+2, slope is -2 and y intercept is 2

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3 years ago

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Gnoma [55]

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3 years ago

Read 2 more answers

.<br><br> 26.<br> Solve 3(2+1) +11 <-2(2 + 13).

ruslelena [56]

Answer:

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3 years ago

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3 years ago

Write a quadratic function with zeroes 0 and 8.

Jlenok [28]

Answer:

$x^2-8x=0$

Step-by-step explanation:

the quadratic function should be as follows:

$x^2-8x=0$

Now let's confirm that the zeros of the function are 0 and 8

$x^2-8x=0=x(x-8)=0$

Therefore we can see that if x = 0

$0^2*8*0=0\\0=0$

the equation is fulfilled

And we also have $(x-8)$

for this expresion to be equal to zero:

$x-8=0\\x=8$

thus, if x = 8

$8^2-8*8=0\\64-64=0\\0=0$

the equation is also fulfilled

The zeros of the quadratic function $x^2-8x=0$ are 0 and 8.

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3 years ago