Ask your own question, for FREE!
Mathematics 22 Online
OpenStudy (anonymous):

PLEASE HELP ME AND WILL GIVE MEDAL! Samsonite16 A student concludes that if x is a real number, then x2 > x. x = -2 x = 3 x = 1 x = -1 Question 4. 4. A student concludes that if x is a real number, then x2 _< x3. Option A: ½ Option B: 0 Option C: 1 Option D:3/2 Question 5. 5. After completing several multiplication problems, a student concludes that the product of two binomials is always a trinomial. (x + 1)(x + 3) (x + 2)(x - 2) (x + 4)(y -3) (x + 5)(x - 6)

OpenStudy (anonymous):

plss hlp

OpenStudy (anonymous):

1st question , is it \[x^{2} > \]

OpenStudy (anonymous):

\[x^{2} > x\]

OpenStudy (anonymous):

or is it just 2x >x

OpenStudy (anonymous):

the 2nd one

OpenStudy (anonymous):

2x>x?

OpenStudy (anonymous):

x^2>x

OpenStudy (anonymous):

then there are two answers when you square a negative number, it becomes a positive. which ones are negative in your answer choices?

OpenStudy (anonymous):

-2 and -1

OpenStudy (anonymous):

(-1)^2 = 1 ****

OpenStudy (anonymous):

(-2)^ 2= 4

OpenStudy (anonymous):

so those two become bigger :) so we can say that x^2 > x for x = -2,-1

OpenStudy (anonymous):

moving on to question 4?

OpenStudy (anonymous):

wait but it has to be one, either -1 or -2 because I dont have -2,-1 as an answer a option

OpenStudy (anonymous):

an answer option***

OpenStudy (anonymous):

Oh! I thought they're asking for what's true :) but in this case, there's another way to get one answer x^2 >x divide x to both sides x>1

OpenStudy (anonymous):

there's only 1 answer that sastisy that condition, which is x=3

OpenStudy (anonymous):

Thank you!

OpenStudy (anonymous):

question 4 now? :)

OpenStudy (anonymous):

Yes please

OpenStudy (anonymous):

Question 5. one of these will disprove the student's conclusion. remember (a-b)(a+b) = a^2 -b^2

OpenStudy (anonymous):

Wait samsonite, are these 3 problems asking us to DISPROVE their answers?

OpenStudy (anonymous):

sorry i was making food. um no i dont think so

OpenStudy (anonymous):

Ok, then those are hints to final answers :)

OpenStudy (anonymous):

Wait whats the answer?

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

Question 4. What is equal to or smaller than 1? In your answer choices. How you do it: first: x^2 _<x^3 divide x^2 to both sides 1 _< x

OpenStudy (anonymous):

what is equal to or LARGER than 1, sorry

OpenStudy (anonymous):

what?

OpenStudy (anonymous):

Hold on sam, I bet your problem includes this portion quote {Choose the counterexample that disproves each conjecture.?)

OpenStudy (anonymous):

was THIS in the direction of the problems you are doing?

OpenStudy (anonymous):

at the beginning when you just start the test , maybe? o.o

OpenStudy (anonymous):

Yeah I think so.

OpenStudy (anonymous):

oh god no wonder why xD I researched a bit on google and found that these questions are asking for the WRONG answer to the problem :)

OpenStudy (anonymous):

so they want the wrong answer to the problem, disproving the problem's equations. the first question would be x = 1 because squares of 1 =1 question 4 would be x = 1/2 as answer because x^2 = 1/4 and x^3= 1/8, 1/4> 1/8, so the students claim is wrong question 5 answer is (x+2)(x-2) because (a-b)(a+b) = a^2 - b^2 which is a binomial :))

OpenStudy (anonymous):

OKay thank you so much!!!! Can I open a new question to give you another medal please for dealing with me. Hahaha xp

OpenStudy (anonymous):

it's ok :) thanks though. You can medal me later if you have another question :))

OpenStudy (anonymous):

Okay thank you again soooo much! And by any chance do you know where to find example powerpoints on CounterExample? @study100

OpenStudy (anonymous):

here :) this video is good, starting at 8:00 minutes (counterexamples) https://www.youtube.com/watch?v=Nu-yNYowcKc#t=496

OpenStudy (anonymous):

i couldn't find his powerpoint though.

OpenStudy (anonymous):

Oh thank you that's great :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!