Question

Can someone help me please

Answer 1

\[2x \sqrt{20x ^{20}}\]

Answer 2

Help me please @satellite73

Answer 3

\[20=4\times 5\] and so \[\sqrt{20}=\sqrt{4\times 5}=\sqrt{4}\sqrt{5}=2\sqrt{5}\] as a start

Answer 4

and also \(\sqrt{x^{20}}=x^{10}\) because \((x^{10})^2=x^{20}\)

Answer 5

so you get
\[2x\sqrt{20x^{20}}=2x\times 2\times x^{10}\sqrt{5}\]

Answer 6

I got \[4x ^{2\sqrt{5x ^{5}}}\]

Answer 7

which you rewrite as
\[4x^{11}\sqrt{5}\]

Answer 8

thanks u so much :)

Answer 9

yw

Answer 10

\[b ^{2}c ^{-1}\] B= -4
C=2

Answer 11

\[(-4)^2\times 2^{-1}\] is the first step

Answer 12

\((-4)^2=(-4)(-4)=16\) and \(2^{-1}=\frac{1}{2}\) so you really have
\[\frac{16}{2}\]

Answer 13

8

Answer 14

better known as 8

Answer 15

\[8\sqrt{x ^{7}}-6\sqrt{x ^{5}}\div2\sqrt{x ^{3}}\]

Answer 16

is it
\[8\sqrt{x^7}-\frac{6\sqrt{x^5}}{2\sqrt{x^3}}\]?

Answer 17

no u dived
all of it by
\[2\sqrt{x ^{3}}\]

Answer 18

\[\frac{8\sqrt{x^7}-6\sqrt{x^5}}{2\sqrt{x^3}}\]

Answer 19

yes

Answer 20

divide each term separately as a start
\[\frac{8\sqrt{x^7}}{2\sqrt{x^3}}-\frac{6\sqrt{x^5}}{2\sqrt{x^3}}\]

Answer 21

you get
\[4\sqrt{x^4}-3\sqrt{x^2}\]because when you divide you subtract the exponents

Answer 22

then \(\sqrt{x^4}=x^2\) and \(\sqrt{x^2}=x\)

Answer 23

okay I get that

Answer 24

so the whole thing is \(4x^2-3x\)

Answer 25

okay that's wat I go t

Answer 26

yay