Need to verify answers

Typaradise

New member
Joined
Feb 3, 2010
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6
1. Square root negative 70 divided by square root negative 7
Answer: square root negative 10

2. negative 9 minus square root negative 18 divided by 3
Answer: negative 3 negative i square root 2
 
Typaradise said:
1. Square root negative 70 divided by square root negative 7

Answer: square root negative 10 No.

\(\displaystyle \frac{\sqrt{-70}}{\sqrt{-7}} \;=\; \sqrt{\frac{-70}{-7}} \;=\; ?\)

2. negative 9 minus square root negative 18 divided by 3 This is poorly worded; what exactly is divided by 3?

Answer: negative 3 negative i square root 2 Did you forget to type the word "minus" ?

I'll guess on this one:


\(\displaystyle \frac{-9 \;-\; \sqrt{-18}}{3} \;=\; \frac{-9 \;-\; 3 \ \sqrt{2} \ i}{3} \;=\; \frac{-9}{3} \;-\; \frac{3 \ \sqrt{2} \ i}{3} \;=\; ?\)

Here's how to type these expressions, using the keyboard.

sqrt(-70) / sqrt(-7)

[-9 - sqrt(-18)] / 3
 
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*1* I think you meant :

\(\displaystyle \frac{\sqrt70}{\sqrt7}\)

Rationalize the denomenator :

\(\displaystyle \frac{\sqrt490}{7} = \frac{7\sqrt10}{7} = \sqrt10\)
 
Aladdin said:
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*1* I think you meant :

\(\displaystyle \frac{\sqrt70}{\sqrt7}\)

Rationalize the denomenator :

\(\displaystyle \frac{\sqrt490}{7} = \frac{7\sqrt10}{7} = /sqrt10\)

Aladdin: In LaTex, if you put curly braces around radicands, you'll get \(\displaystyle \sqrt{70}\) instead of \(\displaystyle \sqrt70\).
 
1)sqrt(-70)/sqrt(-7)=sqrt(10)
2)your answer is right:
(-9-sqrt(18))/3=-3-i*sqrt(2)
 
swaroop said:
2) your answer is right

Swaroop, how can the answer "negative 3 negative i square root 2" be correct, when the answer is "negative 3 minus i square root 2"?
 
Re:

mmm4444bot said:
Aladdin said:
---------------
*1* I think you meant :

\(\displaystyle \frac{\sqrt70}{\sqrt7}\)

Rationalize the denomenator :

\(\displaystyle \frac{\sqrt490}{7} = \frac{7\sqrt10}{7} = /sqrt10\)

Aladdin: In LaTex, if you put curly braces around radicands, you'll get \(\displaystyle \sqrt{70}\) instead of \(\displaystyle \sqrt70\).

Thanks Mark . I'll do this .
 
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