Question

4 over the square root of 21

Answer 1

ok...... so...?

Answer 2

Can you write the equation?

Answer 3

\(\bf \cfrac{4}{\sqrt{21}}\) I'd think, but that means nothing
anymore than just saying.... 27

Answer 4

simplify the radical expression by rationalizing the denominator 4 over the square root of 21

Answer 5

it is the same thing \[4 (21)^{-1/2}\]

Answer 6

But first do the exponent then multiply by 4

Answer 7

just simplify it please

Answer 8

ohhh.. ahemm ok.
well, simplifying a rational with a radical in the denominator means, getting rid of the root at the bottom, and you'd do that by multiplying top and bottom by the denominator, that is \(\bf \cfrac{4}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{4\sqrt{21}}{(\sqrt{21})^2}\implies \cfrac{4\sqrt{21}}{21}\)

Answer 9

and then what do u do after that

Answer 10

well, you can't divide top and bottom because they have no common factors, so that's as much as you'd simplify it

Answer 11

how bout this one

Answer 12

8 over radical 6 - radical 3

Answer 13

do you know what a "conjugate" is?

Answer 14

no

Answer 15

but can you please just solve it for me

Answer 16

please

Answer 17

http://www.mathsisfun.com/algebra/conjugate.html <---- conjugate

gimme a sec, let us use the conjugate of the denominator

Answer 18

i just really need the answer asap

Answer 19

\(\bf{\color{blue}{ (a-b)(a+b) = a^2-b^2}}
\\ \quad \\ \quad \\

\cfrac{8}{\sqrt{6}-\sqrt{3}}\cdot \cfrac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}}\implies \cfrac{8(\sqrt{6}+\sqrt{3})}{(\sqrt{6}-\sqrt{3})(\sqrt{6}+\sqrt{3})}
\\ \quad \\
\cfrac{8(\sqrt{6}+\sqrt{3})}{(\sqrt{6})^2-(\sqrt{3})^2}\)

Answer 20

final answer?

Answer 21

simplify the rational, as you can see the denominator comes out of the radical

Answer 22

ive been trying to do that but cant, please please give me the denominator

Answer 23

plssss

Answer 24

is the denominator 27?????

Answer 25

recall that \(\bf \Large \sqrt{x}\implies \sqrt[{\color{red}{ 2}}]{x^{\color{red}{ 2}}}\implies x\)

Answer 26

??? is the denominator 27?

Answer 27

recall your radicals -> \(\bf \large (\sqrt{6})^2\implies (\sqrt[2]{6})^2\implies \sqrt[2]{6^2}\)

Answer 28

so whats the final answer, just give me that pleaseee

Answer 29

PLEASEEEEE

Answer 30

that removes the neccessity for the exercise

Answer 31

please, im begging you, just give me the answer pleaseeeeeee

Answer 32

just tell me the final answer pleaseeeeeee

Answer 33

ive been stuck for hours

Answer 34

????? please just give me the final answer

Answer 35

.....