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photoshop1234 [79]
3 years ago
11

Write the quotient as a +bi 4+6i/ 2-3i​

Mathematics
2 answers:
vladimir1956 [14]3 years ago
4 0

\frac{4+6i}{2-3i}=\frac{(4+6i)(2+3i)}{(2-3i)(2+3i)}

\frac{8+12i+12i+18i^2}{4+6i-6i-9i^2}

\frac{-10+24i}{13}

\boxed{\frac{-10}{13}+\frac{24}{13}i}

Kipish [7]3 years ago
4 0

Answer:

- \frac{10}{13} + \frac{24}{13} i

Step-by-step explanation:

Given

\frac{4+6i}{2-3i}

To rationalise the denominator, multiply the numerator/ denominator by the complex conjugate of the denominator.

The conjugate of 2 - 3i is 2 + 3i, thus

\frac{(4+6i)(2+3i)}{(2-3i)(2+3i)} ← expand factors

= \frac{8+24i+18i^2}{4-9i^2} → i² = - 1

= \frac{8+24i-18}{4+9}

= \frac{-10+24i}{13}

= - \frac{10}{13} + \frac{24}{13} i

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Answer:

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Step-by-step explanation:

\bf \cfrac{AB}{BC}  =  \tan(60)

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<u>-------------------------------------------</u>

5 0
1 year ago
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Mrs. Robinson, an insurance agent, earns a salary of $4,800 per year plus a 3% commission on her sales. The average price of a p
nekit [7.7K]

Answer:

Required inequality is 4800+183x>8000.

Step-by-step explanation:

Given that Mrs. Robinson, an insurance agent, earns a salary of $4,800 per year plus a 3% commission on her sales. The average price of a policy she sells is $6,100. Write an inequality to find how many policies Mrs. Robinson must sell to make an annual income of at least $8,000.

Calculation is given by:

Salary per year = $4,800

Average price of a policy = $6,100.

commission on her sales = 3%

Then commission on $6,100 = 3% of $6,100 = 0.03 ($6,100) = $183

Let number of policies Mrs. Robinson must sell to make an annual income of at least $8,000 = x

then total commission on x policies = 183x

Total income using x policies = 4800+183x

Since she wants to make an annual income of at least $8,000. so we can write inequality as:

4800+183x>8000

Hence required inequality is 4800+183x>8000.

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ay^2+by+c=a(y-y_1)(y-y_2)

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\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}

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\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}

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