Question

Pam has a complex number with a real part of -2 and an imaginary part of -7. what is the conjugate of pams number? (explain how you got it)

Answer 1

@thomaster @Luigi0210 @preetha @mathmale

Answer 2

Help?

Answer 3

@hartnn

Answer 4

@Compassionate

Answer 5

@agent0smith @mathslover

Answer 6

please help me ASAP i really need help @hartnn @Luigi0210 @thomaster @preetha @mathmale @RadEn @mathslover @Compassionate @agent0smith @Ashleyisakitty

Answer 7

Conjugate of a + bi is a - bi.

Eg. 2 - 4i would have conjugate 2 + 4i

real part of -2 and an imaginary part of -7.
your number is -2 - 7i

Answer 8

i dont understand can you try to futher explain so i can understand

Answer 9

All you do is change the sign (the + or -) on the imaginary part (with the i).

Answer 10

so the answer is simply -2-7i

Answer 11

can you help me with one more problem

Answer 12

show all the work to simplify. leave your answer in simple radical form: 1.7^ sqrt of x^2
2.(^3 sqrt x^4)^8

Answer 13

@agent0smith

Answer 14

@ParthKohli can you help me?

Answer 15

please help walter i need help with the 2nd question

Answer 16

It's really hard to understand the formatting of the second question...

Answer 17

sorry ill try to retype it

Answer 18

\[1.\sqrt[7]{x}^2\] \[2.(\sqrt[3]{x}^4)^8\]

Answer 19

is that good enough @ParthKohli

Answer 20

Ah.

Answer 21

Do you know that\[\large \sqrt[a]{x^b} = x^{b/a}\]

Answer 22

nope

Answer 23

so would number 1 be x^7/2

Answer 24

OK, now you do.

Number one would be\[x^{2/7}\]

Answer 25

is that the final answer?

Answer 26

I believe, yes.

Answer 27

so number 2.

Answer 28

would it be (x^4/3)^8

Answer 29

@ParthKohli ?

Answer 30

Yes!

Answer 31

whats next?

Answer 32

would i evenly distribute the 8

Answer 33

yes, in other words\[(a^m)^n= a^{m \times n}\]

Answer 34

so x^32/24

Answer 35

is that the final answer or would i simplify? @ParthKohli

Answer 36

@ParthKohli

Answer 37

That's not right.

Answer 38

You only multiply to the numerator.

Answer 39

so it would be 768? @parthKohli

Answer 40

Hmm! I must leave now... sorry!

Answer 41

whats the answer

Answer 42

b4 you leave

Answer 43

@ParthKohli please

Answer 44

"so the answer is simply -2-7i"

the conjugate of -2-7i is not -2-7i

Re-read everything I wrote above.

Answer 45

oh im sorry its 2+7i my teacher said something about it im sorry for not putting your answer as a best response and i fanned you thnks

Answer 46

can you actually help me with what my teacher said about my answers@agent0smith

Answer 47

I don't know what your teacher said...

Answer 48

it's not 2+7i either.

Answer 49

Read this carefully:

Conjugate of a + bi is a - bi.

Eg. 2 - 4i would have conjugate: 2 + 4i

your number is -2 - 7i

Find its conjugate.

Answer 50

Here's another example:
-5 + 7i would have conjugate:
-5 - 7i

Answer 51

im sorry my compuyer crashed

Answer 52

wouldnt it be -2+7i then?

Answer 53

Yes

Answer 54

really awesome do you think ytou could help me with some other problems

Answer 55

#1A)
You are correct that you would move the index to the exponent.
However, reverse the fraction. The 2 goes on top, and 7 on the bottom: x^(2/7)

#1B)
Think of it as: x^(4/3)^(8/1)
Then you multiply: 4/3 * 8/1
What do you get for your fraction exponent now?? @agent0smith

Answer 56

Post a new question, this is far too long and is unrelated to the original question

Answer 57

no its just 2 please help i dont want to have to make another post please just help me now and there half don u dont understand certain things in the problems

Answer 58

No, i don't want to look for the original problems in this mess. You'd be better off doing them from the start anyway, than trying to figure out where you went wrong.

Answer 59

but the first one is just \[x^{2/7}\]

Answer 60

what would i do next divide 7 by 2? @agent0smith

Answer 61

You already have the answer

Answer 62

ok so now 2 wouldnt x^(4/3)^(8/1) i multiple the exponents so it would be x^32/3

Answer 63

@agent0smith

Answer 64

@abb0t

Answer 65

yes

Answer 66

i was correct?

Answer 67

yes

Answer 68

well thnks @agent0smith your help was greatly appreciated i fanned you and will prob get around writing a testimonial