Take DEBC a parralelogram
Parralel sides are DE and BC
THEREFORE,
1/2*(a+b)*h
1/2*(6+15)*h
1/2*(21)*h=0
10.5*h=0
h=-10.5
Answer:
B. 1/4
Step-by-step explanation:
Even though A is a fraction it is also an integer (4)
Answer:the mode is b
Step-by-step explanation:
B
Answer:
for this problem you would have to find the coordients on the graph and then once you find all the coordients you would attach the lines together which should be a triangle.
Step-by-step explanation:
If you need answer's I can pin a picture
Let c > 0. Then split the integral at t = c to write
By the FTC, the derivative is
![\displaystyle \frac{df}{dx} = \left(\frac1x + \sin\left(\frac1x\right)\right) \frac{d}{dx}\left[\frac1x\right] - (\ln(x) + \sin(\ln(x))) \frac{d}{dx}\left[\ln(x)\right] \\\\ = -\frac1{x^2} \left(\frac1x + \sin\left(\frac1x\right)\right) - \frac1x (\ln(x) + \sin(\ln(x))) \\\\ = -\frac1{x^3} - \frac{\sin\left(\frac1x\right)}{x^2} - \frac{\ln(x)}x - \frac{\sin(\ln(x))}x \\\\ = -\frac{1 + x\sin\left(\frac1x\right) + x^2\ln(x) + x^2 \sin(\ln(x))}{x^3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdf%7D%7Bdx%7D%20%3D%20%5Cleft%28%5Cfrac1x%20%2B%20%5Csin%5Cleft%28%5Cfrac1x%5Cright%29%5Cright%29%20%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5B%5Cfrac1x%5Cright%5D%20-%20%28%5Cln%28x%29%20%2B%20%5Csin%28%5Cln%28x%29%29%29%20%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5B%5Cln%28x%29%5Cright%5D%20%5C%5C%5C%5C%20%3D%20-%5Cfrac1%7Bx%5E2%7D%20%5Cleft%28%5Cfrac1x%20%2B%20%5Csin%5Cleft%28%5Cfrac1x%5Cright%29%5Cright%29%20-%20%5Cfrac1x%20%28%5Cln%28x%29%20%2B%20%5Csin%28%5Cln%28x%29%29%29%20%5C%5C%5C%5C%20%3D%20-%5Cfrac1%7Bx%5E3%7D%20-%20%5Cfrac%7B%5Csin%5Cleft%28%5Cfrac1x%5Cright%29%7D%7Bx%5E2%7D%20-%20%5Cfrac%7B%5Cln%28x%29%7Dx%20-%20%5Cfrac%7B%5Csin%28%5Cln%28x%29%29%7Dx%20%5C%5C%5C%5C%20%3D%20-%5Cfrac%7B1%20%2B%20x%5Csin%5Cleft%28%5Cfrac1x%5Cright%29%20%2B%20x%5E2%5Cln%28x%29%20%2B%20x%5E2%20%5Csin%28%5Cln%28x%29%29%7D%7Bx%5E3%7D)
