Log(4k - 5) = log(2k - 1)
log(4k) - log(5) = log(2k) - log(1)
0.6020599913k - 0.6989700043 = 0.3010299957k - 0
0.6020599913k - 0.6989700043 = 0.3010299957k
<u>-0.6020599913k -0.6020599913k</u>
<u>-0.6989700043</u> = <u>-0.3010299957k</u>
-0.3010299957 -0.3010299957
2.321928094 = k
The surface area is made up of six sides and the area is then just the sum of the areas of the six sides...
A=2xy+2xz+2yz
A=2(xy+xz+yz) in this case:
A=2(5*4.5+5*3.5+4.5*3.5)
A=2(22.5+17.5+15.75)
A=2(55.75)
A=111.5 in^2
The area of D is given by:
The average value of f over D is given by:
![\frac{1}{ \frac{343}{3} } \int\limits^7_0 \int\limits^{x^2}_0 {4x\sin(y)} \, dydx = -\frac{3}{343} \int\limits^7_0 {4x\cos(x^2)} \, dx \\ \\ =-\frac{3}{343} \int\limits^{49}_0 {2\cos(t)} \, dt=-\frac{6}{343} \left[\sin(t)\right]_0^{49} \, dt=-\frac{6}{343}\sin49](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%20%5Cfrac%7B343%7D%7B3%7D%20%7D%20%20%5Cint%5Climits%5E7_0%20%20%5Cint%5Climits%5E%7Bx%5E2%7D_0%20%7B4x%5Csin%28y%29%7D%20%5C%2C%20dydx%20%20%3D%20-%5Cfrac%7B3%7D%7B343%7D%20%20%5Cint%5Climits%5E7_0%20%7B4x%5Ccos%28x%5E2%29%7D%20%5C%2C%20dx%20%20%5C%5C%20%20%5C%5C%20%3D-%5Cfrac%7B3%7D%7B343%7D%20%5Cint%5Climits%5E%7B49%7D_0%20%7B2%5Ccos%28t%29%7D%20%5C%2C%20dt%3D-%5Cfrac%7B6%7D%7B343%7D%20%5Cleft%5B%5Csin%28t%29%5Cright%5D_0%5E%7B49%7D%20%5C%2C%20dt%3D-%5Cfrac%7B6%7D%7B343%7D%5Csin49)
Answer: 50%
Step-by-step explanation:
There's a 50% chance of landing on tails. So, it would land on tails about 100 times.
<h2>3</h2>
S<--->T
T<--->S
Same thing
<h2>4</h2>
Both sides a equivalent therefore both triangle areas are also equivalent
