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3 years ago

In what quadrant is -14 - 5i located on the complex plane

Mathematics

1 answer:

Mice21 [21]3 years ago

7 0

Put it in standard form, z=a+bi where the point is (a,b)
z=-5i-14
The point would be (-5,-14)
That would be located in the negative x and y coordinates, so it would be in the third or III quadrant.
Hope this helps.

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Solve each equation. Be sure to check for extraneous solution -7= |a-3|/-2

stepan [7]

Answer:

a = -11, 17.

Step-by-step explanation:

-7 = |a-3| / -2

Multiply both sides by -2:

|a - 3| = 14

So a - 3 = 14

Giving a = 17

a - 3 =- -14

giving a = -11.

Checking these values:

-7 = |17-3| / -2 = 14/-2 = -7

- so a = 17 is a solution.

-7 = |-11-3| / -2 = 14/-2 = -7

so a = -11 is also a solution.

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2 years ago

Which expression best estimates. -18 1/4 ÷ 2 2/3 A. 18÷3 B. -18÷3 C.-18÷(-3) D. 18÷(-3)

Volgvan

Answer:

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3 years ago

Complete the definition of the piecewise defined

Westkost [7]

Answer: a=1 b=3

Step-by-step explanation: the x for each of the points are 1 and 3

7 0

2 years ago

A fair coin is tossed three times in succession. The set of equally likely outcomes is StartSet HHH comma HHT comma HTH comma TH

liq [111]

The probability of getting exactly zero tails is 1/8.

Step-by-step explanation:

When a fair coin is tossed three time, the set of equally likely outcomes is:

S= {HHH,HHT, HTH, THH, HTT, THT, TTH, TTT}

n(S)=8

Let A be the event that there are exactly zero tails which means there are all heads

Then

A= {HHH}

n(A)=1

$P(A)=\frac{n(A)}{n(S)}\\=\frac{1}{8}$

The probability of getting exactly zero tails is 1/8.

Keywords: Probability, Equally likely events

Learn more about probability at:

#LearnwithBrainly

5 0

3 years ago

Consider two vectors A and B A=14i and B= -4i+8j

DaniilM [7]

Assuming $\mathbf a,\mathbf b\in\mathbb R^3$ , you have

$\mathbf a\cdot\mathbf b=(14)(-4)+(0)(8)=-56$

so

$((\mathbf a\cdot\mathbf b)\,\mathbf i)\cdot\mathbf a=(-56\,\mathbf i)\cdot(14\,\mathbf i)=(-56)(14)=-784$

Next,

$\mathbf a+\mathbf b=(14\,\mathbf i)+(-4\,\mathbf i+8\,\mathbf j)=10\,\mathbf i+8\,\mathbf j$

Then

$(\mathbf a+\mathbf b)\times\mathbf b=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\10&8&0\\-4&8&0\end{vmatrix}=112\,\mathbf k$

$((\mathbf a+\mathbf b)\times\mathbf b)\times\mathbf k=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&112\\0&0&1\end{vmatrix}=\mathbf 0$

4 0

3 years ago