NobodyAnyone
New member
- Joined
- Dec 13, 2021
- Messages
- 32
I realize that the answer is listed in terms of [imath]\sqrt{x}[/imath], but if y varies directly with x then y = kx.If I understood the question correctly, then the equation has the general form of [imath]y=k\sqrt{x}[/imath].
When x=5 and y = 35, you have [imath]35=k\sqrt{5} [/imath]. Can you solve for k? After that, you can plug in k and x=22 in the general form and solve for y.
For some reason, I skipped the word "directly" entirely. Thanks.I realize that the answer is listed in terms of [imath]\sqrt{x}[/imath], but if y varies directly with x then y = kx.
-Dan
So, why is [imath]\sqrt{x}[/imath] there, and what equation is it saying to solve?I realize that the answer is listed in terms of [imath]\sqrt{x}[/imath], but if y varies directly with x then y = kx.
-Dan
Wdym? I don’t think it lacks anything.So, why is [imath]\sqrt{x}[/imath] there, and what equation is it saying to solve?
I think the question is defective, or at least misleading.
By definition if y varies directly with x then y = kx, not [imath]y = k \sqrt{x}[/imath]. That's what is lacking.Wdym? I don’t think it lacks anything.
Wdym?
Okay then.By definition if y varies directly with x then y = kx, not [imath]y = k \sqrt{x}[/imath]. That's what is lacking.
-Dan
Yeah. My math quizzes always have mistaken in dem.To be crystal clear, there is a typo in the problem and where it says sqrt(x) it should just be x.