Question

@ blacklable would you help me with some quizzes so that i wont get doced?

Answer 1

@BlackLabel i think the first is right the second though im not sure

Answer 2

Sham get out of here its my question

Answer 3

Think of it as this:
\[\huge \sqrt{36 * 2 * z^2}\]
what values can you take out of the sqrt symbol?
babe pls

Answer 4

Okkk ure fine sham
Love you ♥

Answer 5

thanks babe ly2

Answer 6

i got

7 x 2 sqrt2 - 3 x 5 srt 7

Answer 7

6 x 6 = 36

Answer 8

correct, so what is your final answer for the first one?

Answer 9

im on problem 2 i just needed to check if problem 1 was correct

Answer 10

nope, my bad i looked at the wrng one

Answer 11

Shammm ure wrong the first answer is D

Answer 12

i have brainfarts, sometimes

Answer 13

and yeah its d.
\[\huge \sqrt{72z^2} = \sqrt{36*2z^2} = 6z \sqrt{2}\]

Answer 14

Sham and I are getting married ♥

Answer 15

yep we are

Answer 16

oh and back to the question..

Answer 17

congrats to you both :)

Answer 18

and on problem 2 i have 7 x 2 sqrt2 - 3 x 5 srt 7 so would i proceed with the subtraction?

Answer 19

\[\huge 2\sqrt{28} - 5\sqrt{63} = 2\sqrt{7*4} - 5\sqrt{9*7}\]

Answer 20

\[\huge 2*2\sqrt{7} - 5*3\sqrt{7}\]
\[\huge 4\sqrt{7} - 15\sqrt{7}\]

Answer 21

O_OWTF

Answer 22

oh i see i forgot a step its d

Answer 23

becomes
\[\huge -11\sqrt{7}\]

Answer 24

yep

Answer 25

YAYYY sham :*

Answer 26

:*

Answer 27

interesting how this post has turned into a somewhat of a convo

Answer 28

ok so on this one i have to factor the exponents, and ikr

Answer 29

\[\huge \sqrt{20x^3} * \sqrt{32xy^4}\]
\[\huge \sqrt{5*4*x^3} * \sqrt{16*2 *xy^4}\]
simplify it pls

Answer 30

ok so
5x sqrt.4x^2 (x) 4 x y^2sqrt2 ?

Answer 31

and for the second one
if you remember
\[\huge \sqrt[n]{x^m} = x^{\frac{ m }{ n} }\]

Answer 32

the hell did you get that from.

Answer 33

i got it backwards

Answer 34

\[\huge \sqrt{5*4*x^3*16*2*xy^4}\]
\[\huge 16*2 \sqrt{5*2 x^4y^4}\]
\[\huge 8x^2y^2\sqrt{10}\]

Answer 35

that should be a 4*, damn latex

Answer 36

\[2x \sqrt{5x^2} \times 4x^2 \sqrt{2x^2}\]

not sure if this is what you mean

Answer 37

i dont know how you got that answer lol, but what i posted above is the answer for that one

Answer 38

\[\huge (\sqrt[5]{x^6+9})^4 = ((x^6+9)^{\frac{ 1 }{ 5} })^4\]

Answer 39

when you have this
\[\huge (x^m)^n = x^{m*n}\]

Answer 40

option c

Answer 41

so 1/5 x 4 = 4/5
\[\huge (x^6+9)^{\frac{ 4 }{ 5} }\]

Answer 42

yep good job

Answer 43

ok this one if my teacher was correct i just have to factor 729 because of x^7 that would leave me with just x

Answer 44

the sqrt is 27

Answer 45

think of it as this
\[\huge \sqrt[6]{729} * \sqrt[6]{x^7}\]

Answer 46

\[\huge \sqrt[6]{729} = 3 --> 3x \sqrt[6]{x}\]

Answer 47

so thats 3 x \[3 \times \sqrt[6]{x^7}\]

Answer 48

:)

Answer 49

yep

Answer 50

ah i see what you did, mi have no idea why the teacher made it harder than what it really was lol

Answer 51

on the second one you said m / n right?
so
\[\sqrt[4]{(256)^3}\]

Answer 52

yeah

Answer 53

evaluate (256)^3 then take the 4th root

Answer 54

it's the same as
(256)^(3/4)
u could do that too

Answer 55

using a calculator.

Answer 56

oh so :
i got 192 the sqrt of that is a 13.85

Answer 57

(256)^3/4 = 64

Answer 58

you could do it like this
256^3 = 16777216
16777216^(1/4) = 64

Answer 59

i have no idea how to get exponents on this calculator

Answer 60

just use wolfram, makes your life a bit easier

Answer 61

it said 4194304

Answer 62

you probably entered it wrong lol
http://www.wolframalpha.com/input/?i=256%5E%283%2F4%29

Answer 63

lol yes i did i had (256)^3/4

Answer 64

ok so i should just have to guess and check on this one

Answer 65

i think its b or c

Answer 66

okay.
\[\huge -19 * \sqrt[7]{15x+8} = -38\]
\[\huge \sqrt[7]{15x+8} = 2\]
\[\huge (\sqrt[7]{15x+8} )^7 = 2^7\]
\[\huge 15x + 8 = 128\]

Answer 67

\[\huge 15x = 120 , x = 8\]

Answer 68

i thought it was the positive and negative i was supposed to take, how do i check it correctly

Answer 69

what are you talking about

Answer 70

how do i check x = 8 or is there a way to?

Answer 71

i just showed you that x = 8 -_-

Answer 72

you can plug it back into the equation you'll get the same answer

Answer 73

you'll get the answer that it makes the equation true*

Answer 74

i know you showed it just when i plug it back in i don't get the same result

Answer 75

then you probably had an error, with your math

Answer 76

ok

Answer 77

let me try it again

Answer 78

i got it

Answer 79

it is 8

Answer 80

\[\huge -19*\sqrt[7]{15(8) + 8} = -38\]
\[\huge -19 * \sqrt[7]{128} = -38\]
\[\huge -19 * 2 = -38\]
THERFORE i am correct

Answer 81

dunno why u doubt me

Answer 82

i dont doubt you i just was trying to check the answer for my personal knowledge, i see where i had did it wrong...

Answer 83

on the second problem, i know that i is -1

Answer 84

i = √-1
so
\[\huge 2 \sqrt{-27} = 6i \sqrt{3}\]

Answer 85

\[\huge 2\sqrt{-27} ->2\sqrt{9*3*-1} ->2*3*i \sqrt{3}\]
\[\huge 6i \sqrt{3}\]

Answer 86

on this one i had made a simple chart for my class even is negative odd is positive

Answer 87

no it switches at a point

Answer 88

every odd exponent at i is -i

i^2 = -1
i^3 = -i
i^5 = -i
and so on

Answer 89

i^39 = -i
-63i/-9i = 7

Answer 90

the first is working with imaginary numbers so im not sure if i is going to be there

Answer 91

-4 * 10 = -40
i * i = -1
-40 * -1 = 40

Answer 92

anyways i have to go

Answer 93

ok thank you for the help... :)