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3 years ago

(-2i)(5i)(-1) how do you solve by simplifying showing the steps

1 answer:

3 0

$i = \sqrt{-1}$

This is the fundamental property of i that you'll need to keep in mind when solving any problem with imaginary numbers (5i, 2i, 18i, etc.) or complex numbers (3+2i, 8-5i, etc). Squaring both sides of this definition, we also get that

$i^2=-1$

Which we can use to solve this problem. We have:

$(-2i)(5i)(-1)\\
=(-10i^2)(-1)\\
=(-10)(-1)(-1)\\
=-10$

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Identify the solution set of 3 In 4 = 2 In x. -{6} -{-8,8} -(8)

o-na [289]

Option C: The solution set is $\{8\}$

Explanation:

The expression is $3 \ln 4=2 \ln x$

Now, let us find the solution set.

Switch sides, we get,

$2 \ln x=3 \ln 4$

Dividing by 2 on both sides, we have,

$\ln x=\frac{3 \ln 4}{2}$

Thus, $\ln 4=\ln 2^2=2\ln 2$

Hence, the above expression becomes,

$\ln x=\frac{3 \cdot 2 \ln (2)}{2}$

Simplifying, we get,

$\ln x=3 \ln 2$

Applying the log rule, we get,

$$\ln x$=\ln \left(2^{3}\right)$

Simplifying, we have,

$$\ln x$=\ln \left8\right$

Applying the log rule, we have,

$x=8$

Thus, the solution set is $\{8\}$

Hence, Option C is the correct answer.

5 0

3 years ago

Please help it is math

olasank [31]

Answer:

A linear function has the form y=mx+b

A line has a proportional relationship if y/x is always the same ratio for any value.

The slope m=(y2-y1)/(x2-x1) for some two points on a line is always constant, else it wouldn't create a line.

A line won't be proportional if you adapt b because the ratio of y/x won't match the slope anymore.

In the end this means all lines with proportional relationships must intersect (0,0) or in other words f(0)=0.

This happens if they have the shape y=mx.

Step-by-step explanation:

hope this helps pls mark me brainliest

4 0

3 years ago

It x, y, and z are positive intergers and 3x = 4y = 7z, then the least possible value of x + y + z is???

Dmitry [639]

$3x = 4y = 7z \\ \Rightarrow \begin{cases}3x - 4y = 0 \\4y - 7z = 0 \\ 7z - 3x = 0\end{cases} \\ \Leftrightarrow \begin{cases}x = \frac{4}{3} y\\4y - 7z = 0 \\ 7z - 3( \frac{4}{3}y) = 0\end{cases} \\ \Leftrightarrow \begin{cases}x = \frac{4}{3} y\\4y - 7z = 0 \\ 7z - 4y= 0\end{cases} \\ \Leftrightarrow \begin{cases}x = \frac{4}{3} y\\ y \\ z = \frac{4}{7}y \end{cases} \\ x,y,z\: are\: intergers\Rightarrow y=LCM(3,7)=21\Rightarrow x = 28,z = 12 \\ \Rightarrow x + y + z = 61$