Find the constant of variation for the relationship f(x)= 50x. x f(x) 5 50
what will stay the same?
@Bananas1234 ?
f(x) depends on the value 5 and 50 are just a numbers the value of x will never change for the equation
i see
I think that makes sence
so what do you think the answer is?
A?
yes
no
constant of variation is not x
ok
\(\large\color{black}{ \displaystyle y=k~\cdot x }\) is the equation for the direct variation. \(\large\color{black}{ \displaystyle x }\) is the independent variable (it is whatever you plug in for x). \(\large\color{black}{ \displaystyle y }\) is the dependent variable (it depends on what value you are going to plug in for x). \(\large\color{black}{ \displaystyle k }\) is the constant that relates y and x. In this case, whatever you plug it for x, the y will be k times greater than that.
So do i have to slove for K?
You don't need to "solve for k" just compare the (abstract) form y=kx to your equations
equation*
50?
\(\large\color{black}{ \displaystyle y=k~\cdot x }\) formula. \(\large\color{black}{ \displaystyle y=50~\cdot x }\) your's
if you have any questions, please ask.
So if K = 50 would that mean the Constant would be 50?
yes.
Also, "constant" (by definition) is "a number" (not a variable) but any real number.
Thank you. So for future refrence i just compare my given equation to Y = K times X?
well, for this question, or questions that involve different numbers but have the same format, then yes.
Thanks
or, you can say this....
``` "constant of variation" is a number that relates the 2 variables. What are the variables here? Well, they are x and y. What does the 50 do? Any number that you plug in for x (call it a), will be multiplied times 50 to find y (and that is regardless of what the value of a is). So k is relating y and x, by saying that any y-value is greater than x-value [as you plug in any number for x]. ```
hope this made sense just now.
I think so, math is kinda hard for me. I will save that info. thanks.
Alright... not that it is easy for anyone (well, on person's own level). You are welcome!
Join our real-time social learning platform and learn together with your friends!