Question

sr6 x sr5

Answer 1

Hint: use the idea that \[\Large \sqrt{x}\sqrt{y}=\sqrt{xy}\]


In plain English, this says that when you multiply two square roots, you can multiply the stuff inside the square roots and you'll get the same answer


For example \[\Large \sqrt{2}\sqrt{3}=\sqrt{2*3}=\sqrt{6}\]

Answer 2

what if there is exponents for example
3sr2x3sr3

Answer 3

\[\sqrt[3]{2}\sqrt[3]{}\]

Answer 4

oh you mean cubed roots?

Answer 5

yes

Answer 6

same applies


\[\Large \sqrt[3]{x}\sqrt[3]{y} = \sqrt[3]{xy}\]

Answer 7

so for instance,


\[\Large \sqrt[3]{5}\sqrt[3]{6} = \sqrt[3]{5*6}=\sqrt[3]{30}\]

Answer 8

this works for any nth root for that matter

Answer 9

ok thank you, if theres a variable attatched to the sr is it still the same process?

Answer 10

yes this works for any real numbers (and variables are just unknown numbers)

Answer 11

\[\sqrt{7p/6}\sqrt{5/q}\]

Answer 12

same idea also applies to fractions


so multiply the fractions to get


\[\Large \sqrt{\frac{7p}{6}}\sqrt{\frac{5}{q}}=\sqrt{\frac{7p}{6}\times\frac{5}{q}}=\sqrt{\frac{35p}{6q}}\]



So \[\Large \sqrt{\frac{7p}{6}}\sqrt{\frac{5}{q}}=\sqrt{\frac{35p}{6q}}\]

Answer 13

ha its amazing how simple math is once it is properly explained, once again thank you

Answer 14

yw

Answer 15

\[f(x)=\sqrt[3]{40x^6}\]

Answer 16

and what do you want to do here?

Answer 17

oh sorry, find a simplified form of f(x) assume x can be any real number

Answer 18

now you're going in reverse of what you've been doing before


So factor 40 into 8*5 and factor x^6 into x^3*x^3


Doing this gives us


\[\Large \sqrt[3]{40x^6}=\sqrt[3]{8*5*x^3*x^3}\]


Now break up the cubed root


\[\Large \sqrt[3]{40x^6}=\sqrt[3]{8}*\sqrt[3]{5}*\sqrt[3]{x^3}*\sqrt[3]{x^3}\]


and take the cube root of 8 and x^3 to get 2 and x respectively, giving us


\[\Large \sqrt[3]{40x^6}=2*\sqrt[3]{5}*x*x\]


Now just rearrange and multiply the terms to get


\[\LARGE \sqrt[3]{40x^6}=2x^2\sqrt[3]{5}\]

Answer 19

Let me know if you need to me go over that again

Answer 20

oh ok that makes sense.

Answer 21

explain how you could convince a friend that

Answer 22

best way to confirm your answer is to use a graphing calculator

Answer 23

\[\sqrt{x^2-16} greater than or equal \to \sqrt{x^2}-\sqrt{16}\]

Answer 24

if the two expressions make the same exact graph and produce identical tables, then the two expressions are equivalent

Answer 25

oh this is a different problem?

Answer 26

yes sorry i hit post before attaching the problem so confirm your answer to a friend part were the instructions to the problem i just posted

Answer 27

ok

Answer 28

you can also use a graph here

Answer 29

graph the left hand side sqrt(x^2-16) as one equation and sqrt(x^2)-sqrt(16) as another equation

Answer 30

make the first graph a curve you can easily tell the difference from the second

Answer 31

after you've plotted the two graphs together, you'll see that the first curve is always above the second

Answer 32

im not to sure how id go about graphing that

Answer 33

if you have a TI 83 or 84, type into y1 sqrt(x^2-16) and type sqrt(x^2)-sqrt(16) for y2

Answer 34

the sqrt key is on your calculator

Answer 35

got it. much appreciated