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That's the reason I some what dis like High School Math ---- it leaves to many blanks.
When the book writes y = something about x. It is describing a function, some times just a relation, but relations are not as fun.
The proper way to do it, let f be a real map (function) from the real numbers to the real numbers (which I assume you are working on the real numbers), denote f:R ---> R
R stands for real.
Then we let y = f(x) for all x. That's what is going on. y is the out come, or the result of the function performed on x. In this case, we multiply x by an arbitrary k. And that's all there is. It means, y is the result of a function on x that involves multiplying x by a factor k.
Ah hah! We say, one formal definition of a function is its graph, the set of all pairs (x, y) where y = f(x). Are they related here? Yes. It is very important (it's the definition) to link them. What the graph looks like is a straight that is tilted in various degree, 0 degree included to 90 degree excluded. As to WHY it look like that, there is a tool called Calculus, which require some proof to figure. If you want to know, ask or message me.
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Find the variation constant and an equation of variation where y varies directly as x and y equals 10 when x equals 37?
y = kx: 10 = 37k so k = 10/37 and y = 10x/37
What are the values of k when kx plus y equals 4 is tangent to the curve y equals x squared plus 8 showing work?
If: kx+y = 4 and y = x^2+8 Then: x^2+8 = 4-kx => x^2+4+kx = 0 For the line to be tangent to the curve the discriminant of b^2-4(ac) must = 0 So: k^2-4(1*4) = 0 => k^2 -16 = 0 => k^2 = 16 => k = +/- 4 Therefore: y+4x = 4 and y-4x = 4 are tangents to the curve y = x^2+8
What is the value of k when the line y kx plus 1 is a tangent to the curve y2 equals 8x?
If you mean: y = kx +1 and y^2 = 8x So if: y = kx +1 then y^2 = k^2*x^2 +2kx +1 If: y^2 = 8x then k^2*x^2 +2kx +1 = 8x Transposing terms: k^2*x^2 +2kx +1 -8x = 0 Using the discriminant: (2k -8)^2 -4*(k^2*1) = 0 Solving the discriminant: k = 2
Is y equals x over 7 a direct variation?
y=x/7 y=(1/7) x It is a direct variation since it is of the form y=kx, where k=1/7 , a constant
If y varies directly with x and y equals 10 when x equals 5 what is the value of k?
We write y=kx since y varies directly as x. Now we know if x is 5, y is 10. so we write 10=5k so k=2
Find a equation of variation where y varies directly as x and y equals 0.8 when x equals 0.4?
direct variation: y = kx y = kx k = y/x = 0.8/0.4 = 2
A type of variation in the form y equals kx?
Y varies in direct proportion to x.
If y varies directly as x squared and y equals 150 when x equals 5 find y when x equals 4?
First you need to know that the equation you are looking for is y = kx^2. Then you need to substitute the numbers in: y = kx^2 150 = k5^2 150 = k25 6 = k Now that you know k, resubstitute it for the new value of y when x = 4: y = kx^2 y = (6)(4)^2 y = (6)(16) y = 96
How do you Write an equation in the form of y equals kx for the following data?
There are no "following" data!
The variables x and y vary directly when x equals -5 and y equals 21 what is the constant of variation?
y = kx k = y/x = 21/-5 = - 21/5
What is the value of k in the line of y equals kx plus 1 and is tangent to the curve of y squared equals 8x?
If: y = kx+1 is a tangent to the curve y^2 = 8x Then k must equal 2 for the discriminant to equal zero when the given equations are merged together to equal zero.
What if y equals kx?
You then have a linear relationship, or a direct variation. A straight line through the origin.
Suppose y varies directly with x and y equals 12 when x equals 6write an equation realting x and y?
y = kx ie y/x = k, k = 2, so y = 2x
What are the values of k when the line kx plus y equals 4 touches the curve y equals x squared plus 8 on the Cartesian plane showing work?
If: y = x^2 +8 and kx +y = 4 or y = 4 -kx Then: x^2 +8 = 4 -kx So: x^2 +8 -4 +kx = 0 => x^2 +4 +kx = 0 Using the discriminant b^2 -4ac = 0: k^2 -4*1*4 = 0 => k^2 = 16 Therefore the values of k are: -4 or 4 Hence: y-4x = 4 and y+4x = 4 are tangents to the curve y = x^2 +8
What is the value of k when the line y equals kx plus 1.25 is a tangent to the curve y squared equals 10x?
Equations: y = kx +1.25 and y^2 = 10x If: y = kx +1.25 then y^2 = (kx +1.25)^2 =>(kx)^2 +2.5kx +1.5625 So: (kx)^2 +2.5kx +1.5625 = 10x Transposing terms: (kx)^2 +2.5kx +1.5625 -10x = 0 Using the discriminant formula: (2.5k -10)^2 -4(1.5625*k^2) Multiplying out the brackets: 6.25k^2 -50k +100 -6.25^2 = 0 Collecting like terms: -50k +100 = 0 Solving the above equation: k = 2 Therefore the value of k is: 2
If y Varies Directly as x to the 2nd power and y equals 198 when x equals 6 find y when x equals 2?
y=kx^2 hence k=198/36. now y=198/36*(2)^2 y=22
If y varies directly as x and y equals 18 when x equals 3?
y varies directly as x y=kx, where k is a constant y=18 when x=3 18=3k k=6 y=6x
