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2 years ago

Evaluate (2 - 5i)(p + q)i when p = 2 and q = 5ia. 29ib. 291-20C.-21id.29

Mathematics

2 answers:

attashe74 [19]2 years ago

8 0

Answer:

A. 29i

Step-by-step explanation:

Step 1: Plug in given variables

(2 - 5i)(2 + 5i)i

Step 2: Difference of squares (expand)

(4 - 25i²)i

Step 3: Imaginary numbers rules

(4 - 25(-1))i

Step 4: Combine like terms

(4 + 25)i

(29)i

Your final answer will be 29i

kaheart [24]2 years ago

8 0

Answer:

Its A on edg

Step-by-step explanation:

follow my ifunny "dankmemehistory"

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3 years ago

2x-5y=17 ..solve for y

AlladinOne [14]

$2x-5y=17\ \ \ \ | subtract\ 2x\\\\
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3 years ago

Solve: In 2x + In 2 = 0 DONE

pychu [463]

Answer:

$x=\frac{1}{4}$

Step-by-step explanation:

We can use some logarithmic rules to solve this easily.

Note: Ln means $Log_e$

Now, lets start with the equation:

$ln(2x) + ln(2) = 0\\ln(2x) = -ln(2)$

Writing left side with logarithmic base e, we have:

$Log_{e}(2x) = -ln(2)$

We can now use the property shown below to make this into exponential form:

$Log_{a}b=x\\means\\a^x=b$

So, we write:

$Log_{e}(2x) = -ln(2)\\e^{-ln(2)}=2x$

We recognize another property of exponentials:

$a^{bc}=(a^{b})^{c}$

So, we write:

$e^{-ln(2)}=2x\\(e^{ln(2)})^{-1}=2x$

Also, another property of natural logarithms is:

$e^{(ln(a))}=a$

Now, we simplify:

$(e^{ln(2)})^{-1}=2x\\(2)^{-1}=2x\\\frac{1}{2}=2x\\x=\frac{\frac{1}{2}}{2}\\x=\frac{1}{4}$

This is the answer.

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2 years ago

Evaluate (2 - 5i)(p + q)i when p = 2 and q = 5ia. 29ib. 291-20C.-21id.29​

Evaluate (2 - 5i)(p + q)i when p = 2 and q = 5ia. 29ib. 291-20C.-21id.29