Question

Care to help with Liberal Arts Math?

Answer 1

Find the answer to \[\sqrt{169} * \ \frac{ 5 }{ 8}\] is the product nonzero, rational or irrational.

Answer 2

My answer was \[\frac{ 5 }{ 8} \sqrt{13}\] rational

Answer 3

@whpalmer4 Do you think it is correct or should i try again?

Answer 4

That's not correct. Close, but not quite.

Answer 5

what is the square root of 169?

Answer 6

13

Answer 7

so when i multiply 13 and 5/8 i get 8.125...

Answer 8

I would leave it as a fraction, myself, but no matter. your error was that you replaced \(\sqrt{169}\) with \(\sqrt{13}\) instead of \(13\)
\[\sqrt{169}*\frac{5}{8} = 13*\frac{5}{8} = \frac{65}{8} = 8\frac{1}{8}\]

Answer 9

ah! so it would be rational!

Answer 10

and \[8\frac{1}8 = 8.125\]so your calculator work is correct

Answer 11

should I leave it as a fraction?

Answer 12

yes, if you can get rid of the square root sign, it will be rational. if you're stuck with a square root sign, it is irrational.

Answer 13

Care to check my other questions for me? there are not a lot. :)

Answer 14

The problem doesn't say whether you need to express your answer in a particular form, so either way should be okay, but I generally prefer to stay with fractions unless I need decimals. And you could end up with something that doesn't have a fixed length decimal, like 1/3, where writing out the decimal is inconvenient at best...

Answer 15

I could check a few...

Answer 16

Thank you. Okay here is another. it ask to simplify then to write it in radical form
for this I got the answer, \[6x^4\sqrt[11]{y^4}\]

Answer 17

well, first of all, does your assignment say anything about the range of values that \(x\) and \(y\) can have?

Answer 18

No all it ask is to simplify it and then rewrite it in the radical form.

Answer 19

well, it matters because if you don't know what possible values your variables can take on, you can't do some simplifications.

For example, you might think that\[\sqrt{x^2} = x\]right?

Answer 20

You would combine the Y's correct?

Answer 21

It seems like it works...
\[x =3 \]\[x^2 = 9\]\[\sqrt{9} = 3\]

but what if \(x = -1\)?

\[x = -1\]\[x^2 = (-1)(-1) = 1\]\[\sqrt{1} = 1\]\[1 = -1\]uh...not really

Answer 22

Hmm, still confused.

Answer 23

well, probably if you're just starting this, they aren't bothering with such niceties...but strictly speaking, we can't do much simplification on this problem unless we know (or assume) that both \(x\) and \(y\) are not negative.

so, I guess we'll assume that, and you can ask your teacher about whether or not you can simplify if the variables might be negative :-)

Answer 24

let me list the choices.

Answer 25

the original expression to simplify is

\[x^{\frac{5}{3}}\cdot 6\sqrt[3]{x^7}\cdot y^{\frac{3}{4}}\cdot y^2\]did I get that right?

Answer 26

yes it says to simplify then rewrite in radical form

Answer 27

this is application of exponents.

Answer 28

right...can you tell me how to write the term that starts with a 6 as an exponent form, like the other 3 product terms?

Answer 29

Let me list the other choices.

Answer 30

I don't need the other choices...we'll work through this and get the correct answer.

Answer 31

A.\[6x^3\sqrt[9]{x^8}\sqrt{y^3}\]
B.\[6\sqrt[9]{x^35}\sqrt[3]{y^2}\]
C.\[6x^4y^2\sqrt[4]{y^3}\]
D. \[6x^4\sqrt[11]{y^4}\]

Answer 32

Ah sorry lol well working through it I know you need to combine the x and the y values.

Answer 33

once more: can you tell me how to write the term that starts with a 6 as an exponent form, like the other 3 product terms?

Answer 34

ah! you make it \[6x^ \frac{ 7 }{ 3}\]

Answer 35

good. okay, can you combine the two product terms featuring \(x\)?

Answer 36

yes! with the fractions you add them correct?

Answer 37

\[x^a*x^b = x^{a+b}\]

Answer 38

so it becomes \[x^\frac{ 12 }{ 3 }\]

Answer 39

and that becomes 4 correct?

Answer 40

yes. as long as \(x>0\) we can do that simplification, although ideally, we won't lose the \(6\) in the process :-)

Answer 41

so it is 6x^4

Answer 42

now we have Y

Answer 43

so for both of them it would be \[y^ \frac{ 3 }{ 4 } * y ^2\]

Answer 44

yes, and that combines to ...

Answer 45

\[y ^\frac{ 6}{ 4 }\]

Answer 46

or am i wrong?

Answer 47

\[2 + \frac{3}{4} = \frac{6}{4}\]?

Answer 48

hmm..

Answer 49

$2 + $0.75 = $1.50? :-)

Answer 50

OH!

Answer 51

11/4

Answer 52

you multiplied the exponents instead of adding them, I believe...

Answer 53

there you go

how would you write that as a radical expression?

Answer 54

\[\sqrt[11]{y^4}\]

Answer 55

yep, again, assuming that \(y>0\)

so if we combine all the pieces, what is the final answer?

Answer 56

D. aka \[6x^4\sqrt[11]{y^4}\]

Answer 57

Yay so i was right!! Mind if I write this all down ^^

Answer 58

whoops, sorry, that's not right. \[y^{\frac{11}{4}} = \sqrt[4]{y^{11}}\]

Answer 59

but you can simplify that further by factoring out \(y^2\):

\[y^{\frac{11}{4}} = y^{\frac{8}4}*y^{\frac{3}{4}} = y^2*y^{\frac{3}{4}} = y^2\sqrt[4]{y^3}\]

Answer 60

ah okay lemme jot this down

Answer 61

Thank you :)

Answer 62

Now i understand it!!

Answer 63

excellent!

Answer 64

so the correct answer is C, not D

Answer 65

alright sounds good.

Answer 66

Im gonna continue with the others, I will let you know what I got. :)

Answer 67

Okay for this one, it is like the first one, it says to find the answer to 0.2333333...etc. + the square root of 80.
For this i got \[\sqrt{80}\] \[4\sqrt{5}\]

Answer 68

and then for 0.23 i got \[\frac{ 23 }{ 99}\]

Answer 69

now im confused is it rational or irrational

Answer 70

well, first, 0.2333... is not \(23/99\) because \(23/99 = 0.23232323...\)

Answer 71

but if you have radical sign in your expression and cannot simplify it away, you have an irrational number.

Answer 72

Ah okay thank you

Answer 73

\(\sqrt{2}\) irrational
\(\sqrt{3}\) irrational
\(\sqrt[4]\) rational, because you can simplify it to \(2\)
\(\sqrt{5}\) irrational
\(\sqrt{6}\) irrational
etc.

Answer 74

whoops, bumbled the typesetting there, but I think you can figure out what i meant :-)

Answer 75

ah alright, okay for this one, its \[\frac{ 5 }{ 6 }\sqrt{180}\]

Answer 76

I have no idea what to do here.

Answer 77

factor the 180. does it turn into something made up entirely of perfect squares? for example:

\[\sqrt{4} = \sqrt{2*2} = 2\]\[\sqrt{16} = \sqrt{2*2*2*2} = \sqrt{2*2}*\sqrt{2*2} = 2*2 = 4\]

Answer 78

how about 25 and the sqrt of 26

Answer 79

nope that seems to be 130...

Answer 80

\[180 = 2*90 = 2*2*45 = 2*2*3*15 = 2*2*3*3*5 \]right?
so \[sqrt{180} = \sqrt{2*2 * 3*3 * 5} = 2*3*\sqrt{5}\]rational or irrational?

Answer 81

bah, more sloppy typesetting

Answer 82

5(sqrt)5 over 6?

Answer 83

\[\frac{5}{6}\sqrt{180} = \frac{5}{6}*\sqrt{2*2*3*3*5} = \frac{5}{6}*2*3\sqrt{5} = \frac{5}{\cancel{6}}*\cancel{2*3}\sqrt{5} = \]

Answer 84

Still confused. i dont see the sqrt of 5 as the answer...from the choices.

Answer 85

what are the choices? note my answer is not just the square root of 5...

Answer 86

it is \[5\sqrt{5}\]

Answer 87

A. 25(sqrt)25 over 6 irrational
B. 5(sqrt)5 over 6 rational
C.30(sqrt)5 rational
D.5(sqrt)5 irrational

Answer 88

So i was correct?

Answer 89

\[5+\sqrt{98}\] is the next one. let me give you the answer i got.

Answer 90

no, you were not. look at what I did.

\[\frac{5}{6}\sqrt{180} = \frac{5}{6}\sqrt{2*2*3*3*5} = \frac{5}{6}*2*3*\sqrt{5} = \frac{5}{6}*6*\sqrt{5}\]

what is \[\frac{5}{6}*6=\]

Answer 91

just 5

Answer 92

@whpalmer4 that would be 5 correct? so is D the correct answer?

Answer 93

yes, D is the correct answer. and it is irrational because of that \(\sqrt{5}\)

Answer 94

woo! Awesome sorry if Im being a burden. just gets confusing sometimes.

Answer 95

nah, not a burden.

Answer 96

if we do a good job today, that's zillions of questions I don't have to help on in the future, right? :-)

Answer 97

Okay for the one with \[5+\sqrt{98}\]
I seemed to have gotten \[12\sqrt{2}\] irrational
and yep! that is correct!