y=logx becomes 10^y=x
No. It is linear. The giveaway here is the lack of an exponent.
y=4x-8y=2x+10Substitute (4x-8) for y in the second equation and solve for x:4x-8=2x+102x=18x=9Now solve for y:y=4(9)-8=28(9,28)
True
The y-intercept for a pure exponential relationship is always 1.
If the question is, Is y = x4 an exponential function ? then the answer is no.An exponential function is one where the variable appears as an exponent.So, y = 4x is an exponential function.
y=logx becomes 10^y=x
2x - 3y = 142x - y = 10eqn (1) - eqn(2) gives: -2y = 4 so that y = -2Then, by eqn (2): 2x -(-2) = 102x + 2 = 102x = 8 and so x = 4The solution, therefore, is (x, y) = (4, -2)2x - 3y = 142x - y = 10eqn (1) - eqn(2) gives: -2y = 4 so that y = -2Then, by eqn (2): 2x -(-2) = 102x + 2 = 102x = 8 and so x = 4The solution, therefore, is (x, y) = (4, -2)2x - 3y = 142x - y = 10eqn (1) - eqn(2) gives: -2y = 4 so that y = -2Then, by eqn (2): 2x -(-2) = 102x + 2 = 102x = 8 and so x = 4The solution, therefore, is (x, y) = (4, -2)2x - 3y = 142x - y = 10eqn (1) - eqn(2) gives: -2y = 4 so that y = -2Then, by eqn (2): 2x -(-2) = 102x + 2 = 102x = 8 and so x = 4The solution, therefore, is (x, y) = (4, -2)
If your equation is y=0.682x then yes
No. It is linear. The giveaway here is the lack of an exponent.
y=4x-8y=2x+10Substitute (4x-8) for y in the second equation and solve for x:4x-8=2x+102x=18x=9Now solve for y:y=4(9)-8=28(9,28)
Usually expressed as,Y = 3X2============same thing as X * X, XX without multiplicative symbol, so yes it is exponential
yes
True
Since 102x is a multiple of 34x, it is automatically the LCM.
equals(x,y)=1 if x=y =0 otherwise show that this function is primitive recursive
The y-intercept for a pure exponential relationship is always 1.