trustno1
the good girl gone bad
yeah, it'd be cool. if anyone else needs math help(not higher than algebra 2) I'd be happy to help, and I'm sure their are others who would be too
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Anyone here know anything about geometry?
- Thread starter Hero
- Start date
here i go! doing my radical paper! wish me luck 🙂
Double Post (next time use the edit button!):
ok, hit another damn road block.
here's an example.
18. radical 25m to the 9th power
so, then you gotta simplify and ig toes to
radical 25 m8 x radical m
then would the answer be 5m4 radical m or would it be 5m8 radical m?
Double Post (next time use the edit button!):
ok, hit another damn road block.
here's an example.
18. radical 25m to the 9th power
so, then you gotta simplify and ig toes to
radical 25 m8 x radical m
then would the answer be 5m4 radical m or would it be 5m8 radical m?
Ok, it's time we used some brackets. What is to the 9th power? m? Or the whole radical?
Is it: (radical 25m) to the 9th power
or is it: radical 25(m to the 9th power) ?
I'm going to assume for a sec that its the second one. Please note that the carrot symbol (^) means: 'to the power of'. This is an excepted standard for typewritten math. Also, the asterix sumbol (*) is standard instead of a multiplication sign. This is used because (x) is a common algerbreic variable, and it gets confusing.
In this case it becomes
radical (25m^8) * radical m as you already said. Then:
5 * Radical m^8 * radical m
5m^4 * radical m
so your first answer was right.
WARNING: IF THE FOLLOWING CONFUSES YOU JUST IGNORE IT FOR NOW:
________________________
AND, there is a good reason or it. A square root such as radical 25, is no different than 25^1/2, or '25 to the power of 1/2'. Square roots ARE a power of 1/2.
AND Exponents of exponents are multiplied together to get a single exponent.
Eg: M^8^1/2 = M^4.
Eg: X^6^3 = X^18
Eg: A^2^3 = A^6 (because A^2^3 = (AA)^3 = (AA)(AA)(AA) = A^6)
Therefore with radical m^8 (which is really just: 'm to the power of 8 to the power of 1/2'), you multiply the 8 by 1/2 (=4), to get the final power of m^4
Yeah, I know I shouldn't have gone there, but I figured I might as well. Even if it makes no sense now, maybe it will later on. I know I'm a lousy teacher lol. Fortunately we have TrustNo1 around too. 🙂
Is it: (radical 25m) to the 9th power
or is it: radical 25(m to the 9th power) ?
I'm going to assume for a sec that its the second one. Please note that the carrot symbol (^) means: 'to the power of'. This is an excepted standard for typewritten math. Also, the asterix sumbol (*) is standard instead of a multiplication sign. This is used because (x) is a common algerbreic variable, and it gets confusing.
In this case it becomes
radical (25m^8) * radical m as you already said. Then:
5 * Radical m^8 * radical m
5m^4 * radical m
so your first answer was right.
WARNING: IF THE FOLLOWING CONFUSES YOU JUST IGNORE IT FOR NOW:
________________________
AND, there is a good reason or it. A square root such as radical 25, is no different than 25^1/2, or '25 to the power of 1/2'. Square roots ARE a power of 1/2.
AND Exponents of exponents are multiplied together to get a single exponent.
Eg: M^8^1/2 = M^4.
Eg: X^6^3 = X^18
Eg: A^2^3 = A^6 (because A^2^3 = (AA)^3 = (AA)(AA)(AA) = A^6)
Therefore with radical m^8 (which is really just: 'm to the power of 8 to the power of 1/2'), you multiply the 8 by 1/2 (=4), to get the final power of m^4
Yeah, I know I shouldn't have gone there, but I figured I might as well. Even if it makes no sense now, maybe it will later on. I know I'm a lousy teacher lol. Fortunately we have TrustNo1 around too. 🙂
No, atleast I got one right. LoL. I still have about 90 left now. I don't know what I'm going to do, I don't understand anything with people explaining online, going for extra help, having friends telling me, looking at examples. I guess I'm just to stupid... I'll probably give up..... *waits*.... now.
let's see..... about..... 90 questions... and my brother was supposed to help me, but he went out with his friends, so now you guys are my only hope!
so.... would....(radical 18p*4) be.... duh.... um....
18 breaks down to 9 and 2
then 9 breaks down to 3
so it would be 3*3 and 2
would it be 3*3radical2?
so.... would....(radical 18p*4) be.... duh.... um....
18 breaks down to 9 and 2
then 9 breaks down to 3
so it would be 3*3 and 2
would it be 3*3radical2?
Umm.. Radical 18p^4 (p to the power of 4) or Radical 18p*4 (p times 4)?
I'm assuming the first one
So Radical (18p^4)
Radical (9*2p^4) -----------square root of 9 is 3---
3 * Radical (2p^4) ------------Radical (p^4) = p^2----
3p^2 * Radical (2)
Or in english: 3 times (p to the power of 2) times Radical 2.
I'm assuming the first one
So Radical (18p^4)
Radical (9*2p^4) -----------square root of 9 is 3---
3 * Radical (2p^4) ------------Radical (p^4) = p^2----
3p^2 * Radical (2)
Or in english: 3 times (p to the power of 2) times Radical 2.
*slowly reaches for the already loaded gun, and takes the safety off, crying*
ok, um... still not understanding any of this. I got that one righ! except for the extra * thing.... ok... let's try the next one...
(radical 24c7)
so... that goes to....
radical (4*6c^7)
2 * radical 6c^7)
not a clue where to go from there.... how do reduce 6 and 7...
ok, um... still not understanding any of this. I got that one righ! except for the extra * thing.... ok... let's try the next one...
(radical 24c7)
so... that goes to....
radical (4*6c^7)
2 * radical 6c^7)
not a clue where to go from there.... how do reduce 6 and 7...
Ok, you can't reduce the 6, but you can reduce the 7
So you got:
2*radical 6c^7
2 * Radical (6c * c^2 * c^2 * c^2)
2 * Radical (6c) * Radical (c^2) * Radical (c^2) * Radical (c^2)
2 * Radical (6c) * c * c * c
2 * Radical (6c) * c^3
2c^3 * Radical (6c)
like that.
So you got:
2*radical 6c^7
2 * Radical (6c * c^2 * c^2 * c^2)
2 * Radical (6c) * Radical (c^2) * Radical (c^2) * Radical (c^2)
2 * Radical (6c) * c * c * c
2 * Radical (6c) * c^3
2c^3 * Radical (6c)
like that.
Ok here's how.
c to the power of 7 = c^7 = c*c*c*c*c*c*c
so group up the c's by powers of 2
c^2 * c^2 * c^2 * c
You can see that the square root of c^2 = c
Radical (c^2) = c
So if you change
Radical (c^7)
into
Radical (c^2 * c^2 * c^2 *c)
Then you can start taking the c's out of the radical.
Radical (c^2) * Radical (c^2) * radical (c^2) * radical c
= c * c * c * Radical (c)
Hehe sorry I gotta run over to a persons house to pop a network card in - I'll be back a little later to check in.
c to the power of 7 = c^7 = c*c*c*c*c*c*c
so group up the c's by powers of 2
c^2 * c^2 * c^2 * c
You can see that the square root of c^2 = c
Radical (c^2) = c
So if you change
Radical (c^7)
into
Radical (c^2 * c^2 * c^2 *c)
Then you can start taking the c's out of the radical.
Radical (c^2) * Radical (c^2) * radical (c^2) * radical c
= c * c * c * Radical (c)
Hehe sorry I gotta run over to a persons house to pop a network card in - I'll be back a little later to check in.
still struggling over here...
here we go... maybe some examples will help me.
can someone please explain how to do these for me?
1. radical 75q^11
2. radical 12r^6s^7
3. radical 32a^9b^3
4. radical 8bc * radical 4bc
5. radical 6m^4 * radical 3mn^2
6. radical 2a^3 * radical 2c^2x
7. 3 radical 3 * radical 2
8. 3 radical 6 * -4 radical 10
now we got division so im gunna use / for over cuz i dont know the sign
9. 2 radical 7/25
10. 5 radical 9/196
11. 7 radical 11/144
12. 2/radical 3
13. 5/radical 2
14. 2 radical 7/5 radical 3
15. 21/radical 2b
16. radical 21/27d^3
17. radical 10/3 * radical 1/2
18. 5 radical 3 - 2 radical 3 - 6 radical 3
19. 10 radical 8 - radical 72 + 3radical 98
20. radical 2 (radical 3 +3)
21. (8 + radical 7) (8 - radical 7)
any help with any of those would be GREATFULLY appreciated.
here we go... maybe some examples will help me.
can someone please explain how to do these for me?
1. radical 75q^11
2. radical 12r^6s^7
3. radical 32a^9b^3
4. radical 8bc * radical 4bc
5. radical 6m^4 * radical 3mn^2
6. radical 2a^3 * radical 2c^2x
7. 3 radical 3 * radical 2
8. 3 radical 6 * -4 radical 10
now we got division so im gunna use / for over cuz i dont know the sign
9. 2 radical 7/25
10. 5 radical 9/196
11. 7 radical 11/144
12. 2/radical 3
13. 5/radical 2
14. 2 radical 7/5 radical 3
15. 21/radical 2b
16. radical 21/27d^3
17. radical 10/3 * radical 1/2
18. 5 radical 3 - 2 radical 3 - 6 radical 3
19. 10 radical 8 - radical 72 + 3radical 98
20. radical 2 (radical 3 +3)
21. (8 + radical 7) (8 - radical 7)
any help with any of those would be GREATFULLY appreciated.
1:
Radical 75q^11
Radical (3 * 25 * q^11)
5 * Radical (3 * q^2 * q^2 * q^2 * q^2 * q^2 * q)
5 * Radical (3q) * q * q * q * q * q
5 * Radical (3q) * q^5
5q^5 * Radical (3q)
this was the long way of doing it
----------------------------------------------------------
2.
radical 12(r^6)(s^7)
radical [3 * 4(r^6)(s^6)s]
2(r^3)(s^3) * radical (3s)
you can see that when an exponent within the radical is even, dividing it by two will give you the proper exponent outside of the radical. I could have done it the long way:
radical [3 * 4(r^6)(s^7)]
radical (3 * 4 * r * r * r * r * r * r * s * s * s * s * s * s * s)
and then grouped them into perfect squares
radical (3 * 4 * r^2 * r^2 * r^2 * s^2 *s^2 * s^2 * s)
seperated them all into seperate radicals:
radical (3s) * radical (4) * radical (r^2) * radical (r^2) * radical (r^2) * radical (s^2) * radical (s^2) * radical (s^2)
then resolved the perfect squares:
radical (3s) * 2 * r * r * r * s * s * s
Then put all the r's and s's into exponent form
radical (3s) * 2 * r^3 * s^3
2(r^3)(s^3) * radical (3s)
...same answer, but that's the long way, and it's alot of extra writing.
_________________________________________
9:
2 radical (7/25) ------just guessing this is where the parentheses are--------------
2 radical (7) / radical (25)
2 radical (7) / 5
(2/5) radical (7)
The trick in that last one is that whenever you have a fraction in a radical, you can split the radical top and bottom, so radical (7/25) becomes (radical 7) / (radical 25).
----------------------------------------------------------
10:
5 radical (9 / 196)
5 * radical (9) / radical (196)
both of these radicals are perfect squares!
5 * (3 / 14)
15 / 14
1 + 1/14
___________________________________
20.
radical 2 * radical (3 +3) -----did I get the parentheses right?---
radical 2 * radical 6
radical (2 * 6)
radical (12)
radical (3 * 4)
radical (4) * radical (3)
2 radical (3)
----------------------------------------------------------
21. (this one's a little harder)
(8 + radical 7) (8 - radical 7)
FOIL it. It's a binomial, just like in algebra
64 - (8 * radical 7) + (8 * radical 7) - (radical 7)(radical 7)
the middle two cancel out (one's plus, the other minus)
64 - (radical 7)(radical 7)
64 - radical (7 * 7)
64 - radical (49) ---49 is a perfect square----
64 - 7
57
<or you could have just used the Difference of Squares formula, if you remember it from algebra, to finish the question a little quicker, where (a+b)(a-b) = (a^2) - (b^2)>
------------------------------------------------
Does any of this make any sense? I know I'm not a good teacher, so if anyone sees a better way to explain it, please do...
Radical 75q^11
Radical (3 * 25 * q^11)
5 * Radical (3 * q^2 * q^2 * q^2 * q^2 * q^2 * q)
5 * Radical (3q) * q * q * q * q * q
5 * Radical (3q) * q^5
5q^5 * Radical (3q)
this was the long way of doing it
----------------------------------------------------------
2.
radical 12(r^6)(s^7)
radical [3 * 4(r^6)(s^6)s]
2(r^3)(s^3) * radical (3s)
you can see that when an exponent within the radical is even, dividing it by two will give you the proper exponent outside of the radical. I could have done it the long way:
radical [3 * 4(r^6)(s^7)]
radical (3 * 4 * r * r * r * r * r * r * s * s * s * s * s * s * s)
and then grouped them into perfect squares
radical (3 * 4 * r^2 * r^2 * r^2 * s^2 *s^2 * s^2 * s)
seperated them all into seperate radicals:
radical (3s) * radical (4) * radical (r^2) * radical (r^2) * radical (r^2) * radical (s^2) * radical (s^2) * radical (s^2)
then resolved the perfect squares:
radical (3s) * 2 * r * r * r * s * s * s
Then put all the r's and s's into exponent form
radical (3s) * 2 * r^3 * s^3
2(r^3)(s^3) * radical (3s)
...same answer, but that's the long way, and it's alot of extra writing.
_________________________________________
9:
2 radical (7/25) ------just guessing this is where the parentheses are--------------
2 radical (7) / radical (25)
2 radical (7) / 5
(2/5) radical (7)
The trick in that last one is that whenever you have a fraction in a radical, you can split the radical top and bottom, so radical (7/25) becomes (radical 7) / (radical 25).
----------------------------------------------------------
10:
5 radical (9 / 196)
5 * radical (9) / radical (196)
both of these radicals are perfect squares!
5 * (3 / 14)
15 / 14
1 + 1/14
___________________________________
20.
radical 2 * radical (3 +3) -----did I get the parentheses right?---
radical 2 * radical 6
radical (2 * 6)
radical (12)
radical (3 * 4)
radical (4) * radical (3)
2 radical (3)
----------------------------------------------------------
21. (this one's a little harder)
(8 + radical 7) (8 - radical 7)
FOIL it. It's a binomial, just like in algebra
64 - (8 * radical 7) + (8 * radical 7) - (radical 7)(radical 7)
the middle two cancel out (one's plus, the other minus)
64 - (radical 7)(radical 7)
64 - radical (7 * 7)
64 - radical (49) ---49 is a perfect square----
64 - 7
57
<or you could have just used the Difference of Squares formula, if you remember it from algebra, to finish the question a little quicker, where (a+b)(a-b) = (a^2) - (b^2)>
------------------------------------------------
Does any of this make any sense? I know I'm not a good teacher, so if anyone sees a better way to explain it, please do...
trustno1
the good girl gone bad
wow, I never did radicals with powers *looks semi confussed* sorry I can't be much help right now
Originally posted by Khaych
1:
Radical 75q^11
Radical (3 * 25 * q^11)
5 * Radical (3 * q^2 * q^2 * q^2 * q^2 * q^2 * q)
5 * Radical (3q) * q * q * q * q * q
5 * Radical (3q) * q^5
5q^5 * Radical (3q)
this was the long way of doing it
----------------------------------------------------------
2.
radical 12(r^6)(s^7)
radical [3 * 4(r^6)(s^6)s]
2(r^3)(s^3) * radical (3s)
you can see that when an exponent within the radical is even, dividing it by two will give you the proper exponent outside of the radical. I could have done it the long way:
radical [3 * 4(r^6)(s^7)]
radical (3 * 4 * r * r * r * r * r * r * s * s * s * s * s * s * s)
and then grouped them into perfect squares
radical (3 * 4 * r^2 * r^2 * r^2 * s^2 *s^2 * s^2 * s)
seperated them all into seperate radicals:
radical (3s) * radical (4) * radical (r^2) * radical (r^2) * radical (r^2) * radical (s^2) * radical (s^2) * radical (s^2)
then resolved the perfect squares:
radical (3s) * 2 * r * r * r * s * s * s
Then put all the r's and s's into exponent form
radical (3s) * 2 * r^3 * s^3
2(r^3)(s^3) * radical (3s)
...same answer, but that's the long way, and it's alot of extra writing.
_________________________________________
9:
2 radical (7/25) ------just guessing this is where the parentheses are--------------
2 radical (7) / radical (25)
2 radical (7) / 5
(2/5) radical (7)
The trick in that last one is that whenever you have a fraction in a radical, you can split the radical top and bottom, so radical (7/25) becomes (radical 7) / (radical 25).
----------------------------------------------------------
10:
5 radical (9 / 196)
5 * radical (9) / radical (196)
both of these radicals are perfect squares!
5 * (3 / 14)
15 / 14
1 + 1/14
___________________________________
20.
radical 2 * radical (3 +3) -----did I get the parentheses right?---
radical 2 * radical 6
radical (2 * 6)
radical (12)
radical (3 * 4)
radical (4) * radical (3)
2 radical (3)
----------------------------------------------------------
21. (this one's a little harder)
(8 + radical 7) (8 - radical 7)
FOIL it. It's a binomial, just like in algebra
64 - (8 * radical 7) + (8 * radical 7) - (radical 7)(radical 7)
the middle two cancel out (one's plus, the other minus)
64 - (radical 7)(radical 7)
64 - radical (7 * 7)
64 - radical (49) ---49 is a perfect square----
64 - 7
57
<or you could have just used the Difference of Squares formula, if you remember it from algebra, to finish the question a little quicker, where (a+b)(a-b) = (a^2) - (b^2)>
------------------------------------------------
Does any of this make any sense? I know I'm not a good teacher, so if anyone sees a better way to explain it, please do...
Thanks for your help, again. It's really helpful.
12. These are interesting, because all they really want you to do is get the radical off the bottom. To do that we have to get a bit tricky.
2/radical 3
radical (4) / radical (3)
Radical (4/3)
Now here's where we ask: How can I get the 3 off the bottom? Answer: Make the bottom half of the fraction a perfect square. Multiply the top and bottom by whatever the bottom is.
radical (4 * 3 / 3 * 3)
Radical (12/9)
(because 12/9 = 4/3, so we haven't really changed our number)
Radical (12) / Radical (9)
Radical (4 * 3) / 3
[radical (4) * radical (3)] / 3
2/3 radical 3
That should be what they're looking for: Taking the fraction out of the radical.
____________________________________________
18.
5 radical 3 - 2 radical 3 - 6 radical 3
= -3 radical (3)
This is just like if you had this question: 5x - 2x - 6x = -3x
It's no different. Can't reduce it beyond -3 radical (3), so we'll just leave it there.
______________________________________________
19.
10 radical 8 - radical 72 + 3radical 98
Step one, simplify each radical.
[10 radical (4 * 2) ] - radical (36 * 2) + [3 radical (49 * 2)]
Notice that 4, 36 and 49 are perfect squares
[10*2 radical (2)] - [6 radical (2)] + [3 * 7 radical (2)]
20 radical (2) - 6 radical (2) + 21 radical (2)
again, this is just like 20x - 6x + 21x --so solve it that way
= 35 radical (2)
2/radical 3
radical (4) / radical (3)
Radical (4/3)
Now here's where we ask: How can I get the 3 off the bottom? Answer: Make the bottom half of the fraction a perfect square. Multiply the top and bottom by whatever the bottom is.
radical (4 * 3 / 3 * 3)
Radical (12/9)
(because 12/9 = 4/3, so we haven't really changed our number)
Radical (12) / Radical (9)
Radical (4 * 3) / 3
[radical (4) * radical (3)] / 3
2/3 radical 3
That should be what they're looking for: Taking the fraction out of the radical.
____________________________________________
18.
5 radical 3 - 2 radical 3 - 6 radical 3
= -3 radical (3)
This is just like if you had this question: 5x - 2x - 6x = -3x
It's no different. Can't reduce it beyond -3 radical (3), so we'll just leave it there.
______________________________________________
19.
10 radical 8 - radical 72 + 3radical 98
Step one, simplify each radical.
[10 radical (4 * 2) ] - radical (36 * 2) + [3 radical (49 * 2)]
Notice that 4, 36 and 49 are perfect squares
[10*2 radical (2)] - [6 radical (2)] + [3 * 7 radical (2)]
20 radical (2) - 6 radical (2) + 21 radical (2)
again, this is just like 20x - 6x + 21x --so solve it that way
= 35 radical (2)
radical 72c^9d^13
Radical (2 * 36c^9d^13)
36 is a perfect square, so we'll get it out of the way
6 radical (2 c^9 d^13)
Ok, it's important to recognize that the square root of x^2 is x. The square root of x^4 is x^2. The square root of x^6 is x^3. The square root of x^8 is x^4. When you are figuring the square root of a variable with an exponent, just divide the exponent by two. But that works alot better with even numbered exponents, so what we want to do is make those exponents even first: instead of (d^13), we will use (d^12 * d), so that we have an even exponent to work with.
6 radical (2 * c^8 * c * d^12 *d)
6 radical (2) * radical (c^8) * radical (c) * radical (d^12) * radical (d)
now we square root the variables with even numbered exponents, by dividing the exponents by 2, since they ARE perfect squares.
c^8 is a perfect square. The square root of (c^8) = c^4.
d^12 is a perfect square. The square root of (d^12) = d^6.
6 radical (2) * c^4 * radical (c) * d^6 * radical (d)
Answer: 6(c^4)(d^6) radical (2cd)
Radical (2 * 36c^9d^13)
36 is a perfect square, so we'll get it out of the way
6 radical (2 c^9 d^13)
Ok, it's important to recognize that the square root of x^2 is x. The square root of x^4 is x^2. The square root of x^6 is x^3. The square root of x^8 is x^4. When you are figuring the square root of a variable with an exponent, just divide the exponent by two. But that works alot better with even numbered exponents, so what we want to do is make those exponents even first: instead of (d^13), we will use (d^12 * d), so that we have an even exponent to work with.
6 radical (2 * c^8 * c * d^12 *d)
6 radical (2) * radical (c^8) * radical (c) * radical (d^12) * radical (d)
now we square root the variables with even numbered exponents, by dividing the exponents by 2, since they ARE perfect squares.
c^8 is a perfect square. The square root of (c^8) = c^4.
d^12 is a perfect square. The square root of (d^12) = d^6.
6 radical (2) * c^4 * radical (c) * d^6 * radical (d)
Answer: 6(c^4)(d^6) radical (2cd)