(csc x-1)(csc x+1)=csc²x-1²=csc²x-1=(1/sin x)² -1=1/sin x² - sin²x/sin²x=
=(1-sin²x)/sin² x = cos² x/ sin² x = (cos x/sin x)²= (cot x)²
Answer:
x=-21/47
Step-by-step explanation:
Answer:
y = 2*x^2 - 2*x - 24
Step-by-step explanation:
If we have a quadratic function with roots a and b, we can write the equation for that function as:
y = f(x) = A*(x - a)*(x - b)
Where A is the leading coefficient.
In this case, we know that the roots are: 4 and -3
Then the function will be something like:
f(x) = A*(x - 4)*(x - (-3) )
f(x) = A*(x - 4)*(x + 3)
Now we need to determine the value of A.
We also know that the graph of the function passes through the point (3, -12)
This means that:
f(3) = -12
Then:
-12 = A*(3 - 4)*(3 + 3)
-12 = A*(-1)*(6)
-12 = A*(-6)
-12/-6 = A
2 = A
Then the equation is:
y = f(x) = 2*(x - 4)*(x + 3)
Now we need to write this in standard form, so we just need to expand the equation:
y = f(x) = 2*(x^2 + x*3 - x*4 - 4*3)
y = f(x) = 2*(x^2 - x - 12)
y = f(x) = 2*x^2 - 2*x - 24
Then the relation is:
y = 2*x^2 - 2*x - 24
Answer:
Step-by-step explanation:
- If f(x) is in th form of f(x)=g(x)-h(x) then f'(x)=g'(x) - h'(x)
- When f(x)=z(g(x)) then f'(x)= z'(g(x))g'(x) (called as chain rule)
<u>using these information</u>:
g(x)=ln2x then g'(x)=
h(x)=In(3x - 1) then h'(x)=![\frac{(3x-1)'}{3x-1} =\frac{3}{3x-1}f'(x)=g'(x) - h'(x) =[tex]\frac{1}{x} - \frac{3}{3x-1} =\frac{-1}{3x^2-x}](https://tex.z-dn.net/?f=%5Cfrac%7B%283x-1%29%27%7D%7B3x-1%7D%20%3D%5Cfrac%7B3%7D%7B3x-1%7D%3C%2Fp%3E%3Cp%3Ef%27%28x%29%3Dg%27%28x%29%20-%20h%27%28x%29%20%3D%5Btex%5D%5Cfrac%7B1%7D%7Bx%7D%20-%20%5Cfrac%7B3%7D%7B3x-1%7D%20%3D%5Cfrac%7B-1%7D%7B3x%5E2-x%7D)
Answer:
Haha there is no question
Step-by-step explanation:
