Answer:
x=5 and 3
Step-by-step explanation:
x^2-8x+21=6
x^2-8x+21+16=6+16
(x-4)^2=1
(x-4)=±1, x=5 and 3
Zero
There is no solution for these equations as there is no even one point of intersection between them
Answer:
between one and 6
Step-by-step explanation:
musrbe the answer
I'll explain how to do the first one:-
y = cos-1(x2)
This can be described as ' a function of a function' x^2 is a function of x and cos-1(x^2) is a function of x^2.
We need to apply the chain rule.
Personally I find this easier to understand if i let u = x^2, so
If y = f(u) and u is a function of x then
dy/dx = dy/ du * du/dx
Here u = x^2 and y = cos-1(u)
du/dx = 2x
so dy/dx = d(cos-1(x^2) dx = dy/du * du/dx
= -1 / √(1 - u^2) * 2x
= -2x / √(1 - u^2)
= -2x / √(1 - (x^2)^2)
= -2x / √(1 - x^4)
I hope this helps. but if not. you might like to employ the formulae in the question - The square boxes contain the 'u' s in my answer. These formulae are equivalent to my explanation.
3x + 7 > 4x + 3
subtract 3x from both sides
7 > x + 3
7 - 3 > x
x < 4
