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3 years ago

Simplify the expression 2+3i/4+3i Please help

Mathematics

1 answer:

rjkz [21]3 years ago

7 0

Answer:

17+6i

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Step-by-step explanation:

2+3i

---------

4+3i

We want to multiply by the complex conjugate of the denominator

4+3i has a complex conjugate of 4-3i

2+3i 4-3i

--------- * --------------

4+3i 4-3i

Lets multiply the numerator

(2+3i) ( 4-3i)

2*4 +4*3i -3i*2 - 3i*3i =

8 +12i -6i - 9i^2

Remember i^2 = -1

8 +6i -9(-1)

17 +6i

Lets multiply the denominator

(4+3i) ( 4-3i)

4*4 +4*3i -3i*4 - 3i*3i =

16 +12i -12i - 9i^2

Remember i^2 = -1

16 -9(-1)

Putting numerator over denominator

17+6i

-----------

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find the range of the function f(x) = 4x - 1 for the domain {-1, 0, 1, 2, 3}. {-5, -3, 0, 7, 11} {-5, -4, -3, -2, -1} {-11, -7,

Papessa [141]

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3 years ago

I NEED HELP!! please show all work

PtichkaEL [24]

Take the logarithm of both sides. The base of the logarithm doesn't matter.

$4^{5x} = 3^{x-2}$

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Expand the right side:

$\implies 5x \log 4 = x \log 3 - 2 \log 3$

Move the terms containing x to the left side and factor out x :

$\implies 5x \log 4 - x \log 3 = - 2 \log 3$

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Solve for x by dividing boths ides by 5 log(4) - log(3) :

$\implies \boxed{x = -\dfrac{ 2 \log 3 }{ 5 \log 4 - \log 3 }}$

You can stop there, or continue simplifying the solution by using properties of logarithms:

$\implies x = -\dfrac{ \log 3^2 }{ \log 4^5 - \log 3 }$

$\implies x = -\dfrac{ \log 9 }{ \log 1024 - \log 3 }$

$\implies \boxed{x = -\dfrac{ \log 9 }{ \log \frac{1024}3 }}$

You can condense the solution further using the change-of-base identity,

$\implies \boxed{x = -\log_{\frac{1024}3}9}$

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3 years ago

Read 2 more answers

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antiseptic1488 [7]

Answer:

Square each number: 1 , 2 , 3 , 4 , 5:

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3 years ago