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3 years ago

5

Please help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

1 answer:

Alenkasestr [34]3 years ago

6 0

What’s the question?

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A segment in the complex plane has a midpoint at –1 + i. If the segment has an endpoint at –5 – 7i, what is the other endpoint?

saul85 [17]

Answer:

Option D. 3 + 9i

Step-by-step explanation:

We know that the formula to find the middle point is:

$(\frac{x_1 + x_2}{2},\frac{z_1 + z_2}{2})$

We know that the middle point is:

$-1 + i$

And the end point $(x_2, z_2)$ is: $-5 -7i$

We wish to find the point $x_1,z_1$

So we can write the following equation

$\frac{x_1 + (-5)}{2} = -1$ (i)

and

$\frac{z_1 + (-7i)}{2} = i$ (ii)

Now we clear $x_1$ from equation (i) and clear $z_1$ from equation (ii)

For equation (i) we have:

$-2 = x_1 -5 \\\\x_1 = 3$

For equation (ii) we have:

$2i = z_1 -7i\\\\z_1 = 9i$

Then the initial point $x_1, z_1$ is: $3 + 9i$

The correct answer is the last option 3 +9i

3 0

3 years ago

The football team had 72 students try out for the team, which has 3 times as many students who tried out for the volleyball team

Elena-2011 [213]

Answer is 24
72/3 = 24

7 0

3 years ago

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Which of the following equations matches the function shown above

Mnenie [13.5K]

Answer:

where is the function? where are the equations?

7 0

3 years ago

Please answer this correctly without making mistakes

lara31 [8.8K]

Answer: $200 off is the better option.

Step-by-step explanation: 200 is greater than half of 399.80.

3 0

3 years ago

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Let 1 + i be one of the root of z^3 + Az^2 + Bz + 26 = 0, then find the real roots of the equation and determine the value of a

ANEK [815]

Answer:

The real root is -13

The value of A is 11 and B is -24

Step-by-step explanation:

The given equation is $z^3+Az^2+Bz+26=0$ .

If $1+i$ is a root then then the complex conjugate $(1-i)$ is also a root.

Let us now derive the quadratic equation that gave rise to this complex roots.

By the factor theorem, we get;

$(z+(-1-i))(z+(-1+i)=z^2-2z+2$

Now the

Now the fully factored cubic polynomial will be of the form:

$(z+c)(x^2-2x+2)=0$

By comparing constant terms; 2c=26. This means c=13

Hence the required polynomial is $(z+13)(x^2-2x+2)=0$

Therefore the real root is z=-13

We now expand fully to get:

$z^3+11z^2-24z+26=0$

By comparing coefficients; A=11, and B=-24

5 0

4 years ago