Question

SOS I literally have NO idea how to do 9-11!!! If anyone knows how plz help. Thank u so much❤️❤️❤️

Answer 1

Steps:

So before we jump into these problems, we must keep this particular rule in mind:

• Product Rule of Radicals: √ab = √a × √b. Additionally, remember that √a × √a = a

9.

So remember that the perimeter is the sum of all the sides. In this case:

$P=3\sqrt{12} +\sqrt{27} +2\sqrt{48}$

So firstly, using the product rule of radicals we need to simplify these radicals as such:

$3\sqrt{12} =3*\sqrt{4}*\sqrt{3} =3*2*\sqrt{3}=6\sqrt{3}\\\sqrt{27} =\sqrt{9} *\sqrt{3} =3\sqrt{3} \\2\sqrt{48}=2*\sqrt{8}*\sqrt{6}=2*\sqrt{4}*\sqrt{2}*\sqrt{2}*\sqrt{3}=2*2*2*\sqrt{3}=8\sqrt{3}\\\\P=6\sqrt{3}+3\sqrt{3}+8\sqrt{3}$

Now, combine like terms (since they all have the same like term √3, they can all be added up):

$P=6\sqrt{3}+3\sqrt{3}+8\sqrt{3}\\P=17\sqrt{3}$

Your final answer is 17√3 in.

10a.

For this, we will be using the pythagorean theorem, which is $a^2+b^2=c^2$ , where a and b are the legs of the right triangle and c is the hypotenuse of the triangle. In this case, √18 and √32 are our legs and we need to find the hypotenuse. Set up our equation as such:

$(\sqrt{18} )^2+(\sqrt{32})^2=c^2$

From here we can solve for the hypotenuse. Firstly, solve the exponents (remember that square roots and squared power cancel each other out):

$18+32=c^2$

Next, add up the left side:

$50=c^2$

Lastly, square root both sides of the equation:

$\sqrt{50}=c$

The hypotenuse is √50 cm.

10b.

Now, the process is similar to that of 9 so I will just show the steps to the final answer.

$P=\sqrt{18} +\sqrt{32} +\sqrt{50} \\\\\sqrt{18}=\sqrt{9}*\sqrt{2}=3\sqrt{2}\\\sqrt{32}=\sqrt{16}*\sqrt{2}=4\sqrt{2}\\\sqrt{50}=\sqrt{25}*\sqrt{2}=5\sqrt{2}\\\\P=3\sqrt{2}+4\sqrt{2}+5\sqrt{2}\\P=12\sqrt{2}$

The perimeter is 12√2 in.

11.

Now, the process is still similar to question 9 but remember that this time we are working with cube roots.

$P=3\sqrt[3]{24} +\sqrt[3]{27}+2\sqrt[3]{81}\\\\3\sqrt[3]{24}=3*\sqrt[3]{8}*\sqrt[3]{3}=3*2*\sqrt[3]{3}=6\sqrt[3]{3}\\\sqrt[3]{27}=3\\2\sqrt[3]{81}=2*\sqrt[3]{27}*\sqrt[3]{3}=2*3*\sqrt[3]{3}=6\sqrt[3]{3}\\\\P=6\sqrt[3]{3}+3+6\sqrt[3]{3}\\P=3+12\sqrt[3]{3}$

Note that when adding the numbers together, 3 isn't a like term to the other 2 terms because it doesn't have ∛3 multiplied with it.

The perimeter is 3 + 12∛3 in.