Question

What is the simplest form of sqrt75x^2/sqrt12xy^2

Answer 1

@Michele_Laino

Answer 2

These are some more that I am struggling with

Answer 3

starting fro the last:
\[\frac{ 7 }{ 2+\sqrt{5} }=\frac{ 7(2-\sqrt{5}) }{ (2+\sqrt{5})(2-\sqrt{5}) }=\]
=

Answer 4

=\[7\sqrt{5}-14\]

Answer 5

so would that be f?

Answer 6

Use the rule that sqrt(xy)=sqrt(x)*sqrt(y)
sqrt(75x^5)=sqrt(5)*sqrt(5)*sqrt(3)*sqrt(x^4)*sqrt(x).
You can use this to factor out.

Answer 7

now:
\[(2-\sqrt{7})(1+2\sqrt{7})=2+4\sqrt{7}-\sqrt{7}-14=3\sqrt{7}-12\]

Answer 8

@ThePackMan that doesn't make sense

Answer 9

now:
\[2\sqrt{72}-3\sqrt{32}=2\sqrt{9*4*2}-3\sqrt{16*2}=2*3*2\sqrt{2}-3*4\sqrt{2}=0\]

Answer 10

Ok :)

Answer 11

the first one:
\[\frac{ \sqrt{75x ^{5}} }{ \sqrt{12xy ^{2}} }=\]
\[=\frac{ \sqrt{25*3*x ^{4}*x} }{ \sqrt{4*-3*x*y ^{2}} }=\]
\[\frac{ 5x ^{2}\sqrt{x} }{ 2y \sqrt{3x} }=\frac{ 5x ^{2}\sqrt{x} }{ 2\sqrt{3}y \sqrt{x} }=\]
\[\frac{ 5x ^{2} }{ 2y \sqrt{3} }\]

Answer 12

oops: 75=3*25, so your answer is:
\[\frac{ 5x ^{2} }{ 2y }\]

Answer 13

Perfecto! Have time for 4 more?

Answer 14

@Michele_Laino

Answer 15

ok!

Answer 16

Thank you so much! How do I give medals?

Answer 17

from n°8:
\[g(x)-f(x)=x ^{3}-(-2x+5)=x ^{3}+2x-5\]

Answer 18

n°10:
\[\frac{ g(x) }{ f(x) }=\frac{ x+3 }{ x ^{2}-2x-15 }=\]
=\[=\frac{ x+3 }{ (x-5)(x+3) }\]
so domain of g/f= x not equal to 5 and x not equal to -3

Answer 19

n°9:
\[f*g=3x(x ^{2}+1)=3x ^{3}+3x\]

Answer 20

n.7
squaring both sides we have:
\[3x-5=16\]
so 3x=21,
and then x=7 which is acceptable, why?

Answer 21

X=7 because you divided 21 by 3

Answer 22

no, question is more subtle.
hint radicand has to be positive, so?

Answer 23

I have not idea :/