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3 years ago

In which quadrant is the number –14 – 5i located on the complex plane? I II III IV

Mathematics

2 answers:

Dmitry [639]3 years ago

7 0

Answer:

III

Step-by-step explanation:

- 14 tell you that you should be moving left of (0,0) That makes the answer either in quad 2 or quad 3.

- 5i tells you that you are moving down. That tells you that you are moving down.

You are in quad III. It the question was -14 + 5i, you would be moving left and up. That would put you in quad II.

Andrej [43]3 years ago

6 0

Answer:

C. III

Step-by-step explanation:

Edge 2020

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Please help I'm a little confused!

zaharov [31]

I will have "x" represent the unknown number, or you could use a "?", doesn't matter.

$8\leq \frac{x}{-4}$ [8 is less than or equal to (≤) the quotient (÷) of a number and -4, so the number is being divided by -4]

If you need to solve this, you need to isolate/get the variable "x" by itself in the inequality:

$8\leq \frac{x}{-4}$ Multiply -4 on both sides to get rid of the fraction and get "x" by itself

$(-4)8\geq \frac{x}{-4} (-4)$ When you multiply/divide by a negative number, you flip the sign (< >)

-32 ≥ x [-32 is greater than or equal to x, or x is less than or equal to -32]

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3 years ago

1) Find 0.2+(-0.6).<br> » Model the expression on the number line

Temka [501]

Answer:

-0.4

Step-by-step explanation:

If using a number line, start at 0.2 and go backwards 0.6 to get to -0.4

5 0

2 years ago

Will Mark Brainiest!!! Simplify the following:

xz_007 [3.2K]

Answer:

1.B

2.A

3. B

Step-by-step explanation:

1. $\frac{x+5}{x^{2} + 6x +5 }$

We have the denominator of the fraction as following:

$x^{2} + 6x + 5 \\= x^{2} + (1 + 5)x + 5\\= x*x + 1x + 5x + 5*1\\= x ( x + 1) + 5(x + 1)\\= (x + 1) (x + 5)$

As the initial one is a fraction, so that its denominator has to be different from 0.

=> ( $x^{2} +6x+5$ ) ≠ 0

⇔ (x +1) (x +5) ≠ 0

⇔ (x + 1) ≠ 0; (x +5) ≠ 0

⇔ x ≠ -1; x ≠ -5

Replace it into the initial equation, we have:

$\frac{x+5}{x^{2} + 6x +5 } = \frac{x+5}{(x+1)(x+5)}$

As (x+5) ≠ 0; we divide both numerator and denominator of the fraction by (x +5)

=> $\frac{x+5}{x^{2} + 6x +5 } = \frac{x+5}{(x+1)(x+5)} = \frac{1}{x+1}$

So that $\frac{x+5}{x^{2} + 6x +5 } = \frac{1}{x+1}$ with x ≠ 1; x ≠ -5

So that the answer is B.

2. $\frac{(\frac{x^{2} -16 }{x-1} )}{x+4}$

As the initial one is a fraction, so that its denominator has to be different from 0

=> x + 4 ≠ 0

=> x ≠ -4

As $\frac{x^{2}-16 }{x-1}$ is also a fraction, so that its denominator (x-1) has to be different from 0

=> x - 1 ≠ 0

=> x ≠ 1

We have an equation: $x^{2} - y^{2} = (x - y ) (x+y)$

=> $x^{2} - 16 = x^{2} - 4^{2} = (x -4) (x +4)$

Replace it into the initial equation, we have:

$\frac{(\frac{x^{2} -16 }{x-1} )}{x+4} \\= \frac{x^{2} -16 }{x-1} . \frac{1}{x + 4}\\= \frac{(x-4)(x+4)}{x-1}. \frac{1}{x + 4}$

As (x + 4) ≠ 0 (proven above), we can divide both numerator and the denominator of the fraction by (x +4)

=> $\frac{(x-4)(x+4)}{x-1} .\frac{1}{x+4} =\frac{x-4}{x-1}$

So that the initial equation is equal to $\frac{x-4}{x-1}$ with x ≠-4; x ≠1

=> So that the correct answer is A

3. $\frac{x}{4x + x^{2} }$

As the initial one is a fraction, so that its denominator (4x + x^2) has to be different from 0

We have:

(4x + x^2) = 4x + x.x = x ( x + 4)

So that: (4x + x^2) ≠ 0 ⇔ x ( x + 4 ) ≠ 0

⇔ $\left \{ {{x\neq 0} \atop {(x+4)\neq0 }} \right.$ ⇔ $\left \{ {{x\neq 0} \atop {x \neq -4 }} \right.$

As (4x + x^2) = x ( x + 4) , we replace this into the initial fraction and have:

$\frac{x}{4x + x^{2} } = \frac{x}{x(x+4)}$

As x ≠ 0, we can divide both numerator and denominator of the fraction by x and have:

$\frac{x}{x(x+4)} =\frac{x/x}{x(x+4)/x} = \frac{1}{x+4}$

So that $\frac{x}{4x+x^{2} } = \frac{1}{x+4}$ with x ≠ 0; x ≠ -4

=> The correct answer is B

3 0

3 years ago

Write a system of linear equations that has the ordered pair (1,4) as it's solition

Lemur [1.5K]

The required system of equations is $x+y=5\\x-y=-3$

Step-by-step explanation:

We need to write a system of linear equations that has the ordered pair (1,4) as it's solution.

It means we need to find system of linear equations, which after being solved gives x=1 and y=4

Let the system of equations be:

$x+y=5\\x-y=-3$

I have made equations such that adding x+y gives 5 i,e (1+4=5) and subtracting x-y gives -3 (1-4=-3)

Now solving this system of equations to find value of x:

$x+y=5\,\,eq(1)\\x-y=-3\,\,eq(2)$

Adding eq(1) and eq(2)

$x+y=5\\x-y=-3\\-----\\2x=2\\x=1$

Putting value of x=1 into eq(1) to find value of y

$x+y=5\\1+y=5\\y=5-1\\y=4$

The solution set after solving system of equations is (1,4).

The required system of equations is $x+y=5\\x-y=-3$

Keywords: System of equations

Learn more about system of equations at:

#learnwithBrainly

7 0

2 years ago

WILL GIVE BRAINLIEST, SHOW ALL WORK, IF YOU DONT SHOW ALL WORK AND JUST GUESS... I WILL REPORT, SRY BUT PEOPLE JUST WANT POINTS!

Olegator [25]

y = 2x - 3

for x = 0 → y = 2(0) - 3 = -3 → (0, -3)

for x = 2 → y = 2(2) - 3 = 1 → (2, 1)

y = -2x + 5

for x = 0 → y = -2(0) + 5 = 5 → (0, 5)

for x = 2 → y = -2(2) + 5 = 1 → (2, 1)

Answer: A. (2, 1)

6 0

3 years ago