How would the fraction 1/3+square root of 2 be rewritten if its denominator is rationalized using difference of squares? a. 3-square root of 2 b. 3-square root of 2/5 c. 3-square root of 2/7 d. - 3+square root of 2/7

C

given 1 /( 3 + √2 )

then rationalise the denominator by multiplying the numerator/ denominator by the conjugate of the denominator

the conjugate of 3 + √2 is 3 - √2

1 / (3 + √2 ) × (3 - √2 ) / (3 - √2 )

= (3 - √2 ) /( 9 - 2 ) = (3 - √2 ) / 7

is correct

Step-by-step explanation:

the f

last graph i believe

step-by-step explanation:

worked it out i believe

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)