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Aleksandr [31]
4 years ago
10

Joelle was asked to find the missing endpoint of a segment when given an endpoint of –6 +3i and a midpoint of –13 + i. Her work

is shown below.
Her mistake was...


A. not subtracting the a and b in the original equations.

B. writing the final answer as 1+ 5i instead of 5 + i.

C. switching the endpoint and midpoint in her original equations.

D. not using –6 and 3 in one of her original equations and –13 and 1 in the other original equation.

Mathematics
2 answers:
guajiro [1.7K]4 years ago
4 0

Answer:

C. switching the endpoint and midpoint in her original equations.

Step-by-step explanation:

One endpoint is –6 +3i

The other endpoint is a + bi (where a and b are unknown)

The midpoint is -13 + i. Its coordinates are obtained as follows:

coordinate of the midpoint = (coordinate of one endpoint + coordinate of the other endpoint)/2

For this case:

-13 = (-6 + a)/2

1 = (3 +b)/2

shtirl [24]4 years ago
3 0

Answer:

C

Step-by-step explanation:

\frac{-6+a}{2} =-13\\-6+a=-13*2\\a=-26+6\\a=20\\\frac{3i+b i}{2}=i\\\\ \\3+b=2\\b=2-3=-1\\

middle point is 20-i

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Help plzzzz will give brainliest
stich3 [128]

Answer:

It's positive.

Step-by-step explanation:

There are many reasons.... but one of the obvious ones is that it would've shown negative, correct?

8 0
3 years ago
Read 2 more answers
Consider the vectors u <-4,7> and v= <11,-6>
coldgirl [10]

Answer:

u + v = <7 , 1>

║u + v║ ≅ 7

Step-by-step explanation:

* Lets explain how to solve the problem

- We can add two vector by adding their parts

∵ The vector u is <-4 , 7>

∵ The vector v is <11, -6>

∴ The sum of u and v = <-4 , 7> + <11 , -6>

∴ u + v = <-4 + 11 , 7 + -6> = <7 , 1>

∴ The sum u and v is <7 , 1>

* u + v = <7 , 1>

- The magnitude of the resultant vector = √(x² + y²)

∵ x = 7 and y = 1

∵ ║u + v║ means the magnitude of the sum

∴ The magnitude of the resultant vector = √(7² + 1²)

∴ The magnitude of the resultant vector = √(49 + 1) = √50

∴ The magnitude of the resultant vector = √50 = 7.071

* ║u + v║ ≅ 7

4 0
4 years ago
The game of clue involves 6 suspects, 6 weapons, and 9 rooms. one of each is randomly chosen and the object of the game is to gu
mina [271]
Part A:

Given that t<span>he game of clue involves 6 suspects, 6 weapons, and 9 rooms.

The number of ways that one of each is randomly chosen is given by:

^6C_1\times{ ^6C_1}\times{ ^9C_1}=6\times6\times9=324

Therefore, the number of solutions possible is 324.



Part B:

Given that a </span>players is randomly given three of the remaining cards, <span>let s, w, and r be, respectively, the numbers of suspects, weapons, and rooms in the set of three cards given to a specified player.

The number of suspects, weapons, and rooms remaining respectively after the player observes his or her three cards are: 6 - s, 6 - w, and 9 - r.

Let x denote the number of solutions that are possible after that player observes his or her three cards, then:

x={ ^{6-s}C_1}\times{ ^{6-w}C_1}\times{ ^{9-r}C_1}=(6-s)(6-w)(9-r)

Therefore, x in terms of s, w, and r is given by x = (6 - s)(6 - w)(9 - r).



Part C:

The expected value E(x) of a data set x_i with probabilities p(x_i) is given by E(x)=\Sigma xp(x)

There are </span>^{3+3-1}C_{3-1}={ ^5C_2}=10 possible combinations s, w and r. They are (3, 0, 0), (0, 3, 0), (0, 0, 3), (2, 1, 0), (0, 2, 1), (1, 0, 2), (2, 0, 1), (1, 2, 0), (0, 1, 2), (1, 1, 1)

Thus the expected value is given by

E(x)=3\cdot6\cdot9p(3, 0, 0)+6\cdot3\cdot9p(0, 3, 0)+6\cdot6\cdot6p(0, 0, 3) \\ 4\cdot5\cdot9p(2, 1, 0)+6\cdot4\cdot8p(0, 2, 1)+5\cdot6\cdot7p(1, 0, 2)+4\cdot6\cdot8p(2, 0, 1) \\ +5\cdot4\cdot9p(1, 2, 0)+6\cdot5\cdot7p(0, 1, 2)+5\cdot5\cdot8(1, 1, 1) \\  \\ = \frac{1}{ ^{21}C_3} (162\cdot{ ^6C_3}\cdot{ ^6C_0}\cdot{ ^9C_0}+162\cdot{ ^6C_0}\cdot{ ^6C_3}\cdot{ ^9C_0}+216\cdot{ ^6C_0}\cdot{ ^6C_0}\cdot{ ^9C_3} \\ \\ +180\cdot{ ^6C_2}\cdot{ ^6C_1}\cdot{ ^9C_0}+192\cdot{ ^6C_0}\cdot{ ^6C_2}\cdot{ ^9C_1}

+210\cdot{ ^6C_1}\cdot{ ^6C_0}\cdot{ ^9C_2}+192\cdot{ ^6C_2}\cdot{ ^6C_0}\cdot{ ^9C_1}+180\cdot{ ^6C_1}\cdot{ ^6C_2}\cdot{ ^9C_0} \\  \\ +210\cdot{ ^6C_0}\cdot{ ^6C_1}\cdot{ ^9C_2}+200\cdot{ ^6C_1}\cdot{ ^6C_1}\cdot{ ^9C_1} \\  \\ =\frac{1}{1,330}(324\cdot20+216\cdot84+360\cdot90+384\cdot135+420\cdot216+200\cdot324) \\  \\ =\frac{1}{1,330}(6,480+18,144+32,400+51,840+90,720+64,800) \\  \\ =\frac{1}{1,330}(264,384) \\  \\ =\bold{198.78}
6 0
3 years ago
Is 67.714813 rational or irrational?
Komok [63]
Hi,
Answer is:
Irrational. Why?

Rational numbers are like 2:3 (Example) that is a Decimal. 

Hope this helps!! Can you please mark Brainliest? Good Luck on your School assignment, and God Bless!


4 0
3 years ago
HELP ASAP PLEASE AND SHOW WORK!!!
Valentin [98]

Answer:

<h3>             f(x) = 2(x - 6)² - 3 </h3>

Step-by-step explanation:

f(x) = a(x - h)² + k   ← vertex form of parabola equation with vertex (h, k)

So:

f(x) = a(x - 6)² + (-3)

f(x) = a(x - 6)² - 3      ←  vertex form of our parabola equation

Parabola goes through (8, 5) so if x=8 then f(x)=5

5 = a(8-6)² - 3

5 +3 = a(2)² - 3  +3

8 = 4a

a = 2

That means, the equation of a parabola with vertex (6, -3) and passing through the point (8, 5):

                                         f(x) = 2(x - 6)² - 3

4 0
3 years ago
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