Tutors Answer Your Questions about Radicals (FREE)
Question 256858: OK I have a problem: I have the right answer and I've even had it double checked by my teacher I'm just not sure how I got there.
The problem is: 2√5+3√20 And I know the right answer is:8√5.
I just can't remember I got from one to the other.
Click here to see answer by CharlesG2(828) |
Question 256858: OK I have a problem: I have the right answer and I've even had it double checked by my teacher I'm just not sure how I got there.
The problem is: 2√5+3√20 And I know the right answer is:8√5.
I just can't remember I got from one to the other.
Click here to see answer by Earlsdon(6294) |
Question 256858: OK I have a problem: I have the right answer and I've even had it double checked by my teacher I'm just not sure how I got there.
The problem is: 2√5+3√20 And I know the right answer is:8√5.
I just can't remember I got from one to the other.
Click here to see answer by richwmiller(9144)  |
Question 257246: I have gotten all my questions right in the Lesson except for this one.I'm not sure why this one seems so hard to me, I'm not even sure how to start. Any help is greatly appreciated.
The Problem: (-3125a5)^(2/5)
Click here to see answer by Alan3354(31534)  |
Question 258800: Hi
Simplify the radical by rationalizing the denominator. Variables are assumed to be positive.
square root sign over the following: 3x to the 6th power over 2y to the 4th power. The square root sign is over the whole fraction with a small 5 to the left of the square root sign. All to the 5th power.
I'm having trouble figuring out what to do.
Thanks
R
Click here to see answer by stanbon(57979) |
Question 259110: I hope I found the right section to submit this to.
the question is
J=kGV
J=the square root of 3
G= the square root of 7
V= the square root of 5
what is the constant variant k?
It is supposed to equal 3 square roots of 5. I cannot figure out how they got 3 square roots of 5
Click here to see answer by richwmiller(9144)  |
Question 259158: This formula is supposed to be carburetor mm's open, and the resulting rpm's for peak performance with 2.8L displacement. Please advise on calculations of this formula:
D(p)=.75√(2.8)p
D=81
p=4166
If you would, please help with calculations to arrive at 4166, and I also need to reverse the formula where D=84, and p=4480. I'm not sure how to do this either.
Thank you!
dmaher
Click here to see answer by Fombitz(13828)  |
Question 259287: Please help with an algebra application with carburetor open mm's and rpm's. I need to reverse this formula with the existing p=4480 and D=84 for the following, and instead of p=4480 as the result, need D=84 as the result.
Thank you very much!
D(p)=0.75√2.8p
D(p)=(3/4)(√14/5)p
D*2=(9/16)(14/5)p
p=(16/9)(5/14)D^2
p=(16/9)(5/14)(84)^2
p=4480
Click here to see answer by ankor@dixie-net.com(15746)  |
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