Step-by-step explanation:
We can prove the statement is false by proof of contradiction:
We know that cos0° = 1 and cos90° = 0.
Let A = 0° and B = 90°.
Left-Hand Side:
cos(A + B) = cos(0° + 90°) = cos90° = 0.
Right-Hand Side:
cos(A) + cos(B) = cos(0°) + cos(90°)
= 1 + 0 = 1.
Since LHS =/= RHS, by proof of contradiction,
the statement is false.
The hyperbolic cos (cosh) is given by
cosh (x) = (e^x + e^-x) / 2
The slope of a tangent line to a function at a point is given by the derivative of that function at that point.
d/dx [cosh(x)] = d/dx[(e^x + e^-x) / 2] = (e^x - e^-x) / 2 = sinh(x)
Given that the slope is 2, thus
sinh(x) = 2
x = sinh^-1 (2) = 1.444
Therefore, the curve of y = cosh(x) has a slope of 2 at point x = 1.44
Answer:
Step-by-step explanation:
I would start by visually thinking out this problem. Or if you have time sketch it out. And don't forget the popcorn :)
Then while doing this retry the problem
This is called the <u>Multiplication property of equality.</u>
The multiplication property of equality states that whatever you multiply on one side, you must multiply the same quantity on the other side of the equation. In this case, variable c is multiplied to both sides of the equation.