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3 years ago

Help me pleaseeeeeeeeeeeeeeeee

Mathematics

1 answer:

hjlf3 years ago

3 0

Answer:

The Order of operations is PEMDAS: Parentheces, Exponents, Multiplcation, Division, Addition, Subtraction

Step-by-step explanation:

$9+3*2-(6+3^2) \\(6+3^2= 12\\3*2= 6\\9+6=15\\\\15-12= 3$

Option A is not correct because it equals 9.

Option B is not correct because it equals -66

Option C is CORRECT because it equals 3 that is your ANSWER

Option D is not correct because it equals 0

PLZ PLZ PLZZZZ MARK BRAINLIEST I WORKED VERY HARD ON THIS

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3 years ago

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Step-by-step explanation:

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3 years ago

Find the multiplicative inverse of 6 + 2i

Marina CMI [18]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2774989

________________

Find the multiplicative inverse of

$\mathsf{z=6+2i}$

________

The inverse multiplicative of $\mathsf{z=a+bi}$ is

$\mathsf{\dfrac{1}{z}}\\\\\\
=\mathsf{\dfrac{1}{a+bi}\qquad\quad(a\ne 0~~and~~b\ne 0)}\\\\\\
=\mathsf{\dfrac{1}{a+bi}\cdot \dfrac{a-bi}{a-bi}}\\\\\\
=\mathsf{\dfrac{1\cdot (a-bi)}{(a+bi)\cdot (a-bi)}}\\\\\\
=\mathsf{\dfrac{a-bi}{a^2-\,\diagup\hspace{-10}abi+\,\diagup\hspace{-10}abi-(bi)^2}}$

$=\mathsf{\dfrac{a-bi}{a^2-b^2\cdot i^2}}\\\\\\
=\mathsf{\dfrac{a-bi}{a^2-b^2\cdot (-1)}}\\\\\\
=\mathsf{\dfrac{a-bi}{a^2+b^2}}\\\\\\\\
\therefore~~\mathsf{\dfrac{1}{a+bi}=\dfrac{a}{a^2+b^2}-\dfrac{b}{a^2+b^2}\,i\qquad\quad\checkmark}$

________

For this question,

$\mathsf{z=6+2i}$

So,

$\mathsf{\dfrac{1}{z}}\\\\\\
=\mathsf{\dfrac{1}{6+2i}}\\\\\\
=\mathsf{\dfrac{6}{6^2+2^2}-\dfrac{2}{6^2+2^2}\,i}\\\\\\
=\mathsf{\dfrac{6}{36+4}-\dfrac{2}{36+4}\,i}\\\\\\
=\mathsf{\dfrac{6}{40}-\dfrac{2}{40}\,i}$

$\therefore~~\mathsf{\dfrac{1}{z}=\dfrac{3}{20}-\dfrac{1}{20}\,i}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

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3 years ago

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