Answer:
Step-by-step explanation:
Well lets start with
(3*sqrt 2+ 1)(2*sqrt 3-4)
Lets multiply everything in the second parenthesis by 3*sqrt 2
2 sqrt 3 * 3 sqrt 2 = 6 sqrt 6
-4*3 sqrt 2 = -12 sqrt 2
Now lets multiply everything by 1
1*2 sqrt 3 = 2 sqrt 3
1*-4 = -4
we have
-4 -12 [tex]\sqrt{2}[/tex] + 6 [tex]\sqrt{6}[/tex] + 2 [tex]\sqrt{3}[/tex] as the awnser to problem A
Now lets do problem b
We can start by multiplying everything in the second parenthesis by 2
2*5=10
2*-1 *sqrt 3 = -2 sqrt 3
Now multiply everything in the second parenthesis by 2sqrt3
2sqrt 3 * 5 = 10 sqrt 3
2 sqrt 3 * -1* sqrt3 = -6
Our final awnser is
10-6 +10 sqrt 3 -2 sqrt 3 -> 4+ 8 [tex]\sqrt{3}[/tex]
The awnser to question B is 4+ 8 [tex]\sqrt{3}[/tex]
pls give brainliest