Question

please help me with algebra 2?! i am so lost on how to do this!
1.Multiply and simplify if possible.

four square root of eleven•four square root of ten
(1 point)

11four square root of ten
11
10
four square root of one hundred ten

2.What is the simplest form of the expression?

cubed root of twenty four times a to the tenth times b to the sixth
(1 point)

3a3b2cubed root of two a
2a3bcubed root of a
2a3b2cubed root of three a
none of these

Answer 1

\[4\sqrt{11}\times 4\sqrt{10}\] like that?

Answer 2

or maybe
\[\sqrt[4]{11}\times \sqrt[4]{10}\]

Answer 3

if it is the second one, that does not read "four square root" that reads "the fourth root" and you would multiply inside the radical to get
\[\sqrt[4]{10}\sqrt[4]{11}=\sqrt[4]{110}\]

Answer 4

yes the second one is what the problem looks like. can u u help me?

Answer 5

yeah if the "index" is the same ( the index in your example is 4) you can multiply inside the racial, as i wrote above

Answer 6

since you have no idea what number raised to the fourth power gives 110, just leave it as \(\sqrt[4]{110}\)

Answer 7

\[\sqrt[3]{24a^{10}b^6}\] is a little different

Answer 8

ok i got the first one. its was as hard as i was making it out to be. now how do i go the next one?

Answer 9

since \(24=3\times 8\) and since \(2^3=8\) we can write
\[\sqrt[3]{24}=\sqrt[3]{8\times 3}=\sqrt[3]{8}\sqrt[3]{3}=2\sqrt[3]{3}\]

Answer 10

as for \(\sqrt[3]{b^6}\) since 3 goes in to 6 twice, that is the same as \(b^2\)

Answer 11

and for \(\sqrt[3]{a^{10}}\) since 3 goes in to ten 3 times with a remainder of 1, that is the same as \(a^3\sqrt[3]{a}\)

Answer 12

all together that makes
\[\sqrt[3]{24a^{10}b^6}=2a^3b^2\sqrt[3]{3a}\]

Answer 13

ok i think i get what your doing.

Answer 14

3.
What is the simplest form of the product?\[^3\sqrt{4x^2} \times ^3\sqrt{8x^7}\]

Answer 15

all hail @satellite73 the dark lord and savior

Answer 16

#3 is \[2x^{3} \times \sqrt[3]4\]