A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing

This Question is Closed

Lynncake
 one year ago
Best ResponseYou've already chosen the best response.0Question 7: A \[(8+\sqrt{2})(43\sqrt{2}) \rightarrow 32 24\sqrt{2}+4\sqrt{2}3\sqrt{2}\sqrt{2}\]\[3220\sqrt{2}3(2) \rightarrow 32620\sqrt{2} \rightarrow 2620\sqrt{2}\]

AriPotta
 one year ago
Best ResponseYou've already chosen the best response.1and with the first one, just multiply the two fractions. 2/3 x 4/5 = 8/15

miah
 one year ago
Best ResponseYou've already chosen the best response.0Which of the following is a solution to √x4+9=6

miah
 one year ago
Best ResponseYou've already chosen the best response.0x = 17 x = 15 x = 5 x = 3

AriPotta
 one year ago
Best ResponseYou've already chosen the best response.1is that sqrt(x  4) + 9 = 6?

miah
 one year ago
Best ResponseYou've already chosen the best response.0x = –3 x = –7 x = 1 x = 13 I put the wrong answer choices

AriPotta
 one year ago
Best ResponseYou've already chosen the best response.1ok, so subtract 9 first

AriPotta
 one year ago
Best ResponseYou've already chosen the best response.1sqrt(x  4) + 9 = 6 sqrt(x  4) = 3 then square it [sqrt(x  4)]^2 = 3^2 x  4 = 9 add 4 x = 13

miah
 one year ago
Best ResponseYou've already chosen the best response.0i have one more. I gonna see if I can do it

miah
 one year ago
Best ResponseYou've already chosen the best response.0x = 17 x = 15 x = 5 x = 3

AriPotta
 one year ago
Best ResponseYou've already chosen the best response.1divide by 6 6sqrt(x + 1) = 24 sqrt(x + 1) = 4 square it [sqrt(x + 1)]^2 = 4^2 x + 1 = 16 subtract 1 x + 1 = 16 x = 15
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.