Answer:
-3
Step-by-step explanation:
9×(12+(-5)÷(-8-13)
9×7÷(-8-13)
63÷-21
=-3
cos θ = Adjacent/ hypotenuse
cosθ= 5/13
a²+b²= c²
a² + 5² = 13²
a² = 13² - 5²
a² = 144
a=√144
a= 12
<u>a</u> is the opposite = 12
<u>b</u> is the Adjacent = 5
<u>c</u> is the hypotenuse = 13
a) tan θ= opposite/Adjacent
tan θ = 12/5
b) sin θ= opposite/ hypotenuse
sinθ= 12/13
C) sec θ= hypotenuse / Adjacent
sec θ= 13/5
d) cscθ= hypotenuse /opposite
cscθ= 13/12
e) cotθ=Adjacent/ opposite
cotθ= 5/12
Answer:
f'(x) > 0 on
and f'(x)<0 on
Step-by-step explanation:
1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:

To find its decreasing interval :

2) Then let's find the critical point of this function:
![f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0](https://tex.z-dn.net/?f=f%27%28x%29%3D%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cmathrm%7Bd%7D%20x%7D%5B6-2%5E%7B2x%7D%5D%3D%5Cfrac%7B%5Cmathrm%7Bd%7D%20%7D%7B%5Cmathrm%7Bd%7Dx%7D%5B6%5D-%5Cfrac%7B%5Cmathrm%7Bd%7D%7D%7B%5Cmathrm%7Bd%7Dx%7D%5B2%5E%7B2x%7D%5D%3D0-%5Bln%282%29%2A2%5E%7B2x%7D%2A%5Cfrac%7B%5Cmathrm%7Bd%7D%7D%7B%5Cmathrm%7Bd%7Dx%7D%5B2x%5D%3D-ln%282%29%2A2%5E%7B2x%7D%2A2%3D-ln2%2A2%5E%7B2x%2B1%5CRightarrow%20%7Df%27%28x%29%3D-ln%282%29%2A2%5E%7B2x%7D%2A2%5C%5C-ln%282%29%2A2%5E%7B2x%2B1%7D%3D-2x%5E%7B2x%7D%28ln%28x%29%2B1%29%3D0)
2.2 Solving for x this equation, this will lead us to one critical point since x' is not defined for Real set, and x''
≈0.37 for e≈2.72

3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.
Answer I believe that the answer is 1/36.
Step-by-step explanation: I just toke the test and got it right.
Answer:
x1= -9/2 ; x2= -7/3
Step-by-step explanation:
Rewrite to 6x^2 + 27x +14x +63 = 0
Factor and get 3x(2x + 9) + 7(2x + 9) = 0
Factor again and get (2x + 9)(3x + 7) = 0
Separate into possible cases -> 2x + 9 = 0; 3x + 7 = 0
The equation has two solutions: x1 = -9/2 and x2 = -7/3
Hope this helped!!