QUESTION 1
The given logarithm is
We apply the power rule of logarithms; 
We now apply the product rule of logarithm;
QUESTION 2
The given logarithm is
We apply the power rule of logarithm to get;
We apply the product to obtain;
We apply the quotient rule; 
![=\log_5(\frac{x^8 \sqrt[4]{y^3} }{z^5})](https://tex.z-dn.net/?f=%3D%5Clog_5%28%5Cfrac%7Bx%5E8%20%5Csqrt%5B4%5D%7By%5E3%7D%20%7D%7Bz%5E5%7D%29)
270,050 because you just add a zero
The answer is 2.58 moles potassium<span>
The solution to this is,
to find moles, convert grams to mole by the equation
grams(mole/gram)=moles
101. g of potassium(1 mole potassium/</span><span>39.10 g potassium)=2.58 moles potassium</span>
From the figure shown, the interval is divided into 5 equal parts making each subinterval to be 0.2.
Part A:
The approximate the area of the region shown in the figure using the lower sums is given by:
![Area= [y(0.2)\times0.2]+[y(0.4)\times0.2]+[y(0.6)\times0.2]+[y(0.8)\times0.2] \\ +[y(1)\times0.2] \\ \\ =[\sqrt{1-(0.2)^2}\times0.2]+[\sqrt{1-(0.4)^2}\times0.2]+[\sqrt{1-(0.6)^2}\times0.2] \\ +[\sqrt{1-(0.8)^2}\times0.2]+[\sqrt{1-(1)^2}\times0.2] \\ \\ =(0.9798\times0.2)+(0.9165\times0.2)+(0.8\times0.2)+(0.6\times0.2)+(0\times0.2) \\ \\ =0.196+0.183+0.16+0.12=0.659](https://tex.z-dn.net/?f=Area%3D%20%5By%280.2%29%5Ctimes0.2%5D%2B%5By%280.4%29%5Ctimes0.2%5D%2B%5By%280.6%29%5Ctimes0.2%5D%2B%5By%280.8%29%5Ctimes0.2%5D%20%5C%5C%20%2B%5By%281%29%5Ctimes0.2%5D%20%5C%5C%20%20%5C%5C%20%3D%5B%5Csqrt%7B1-%280.2%29%5E2%7D%5Ctimes0.2%5D%2B%5B%5Csqrt%7B1-%280.4%29%5E2%7D%5Ctimes0.2%5D%2B%5B%5Csqrt%7B1-%280.6%29%5E2%7D%5Ctimes0.2%5D%20%5C%5C%20%2B%5B%5Csqrt%7B1-%280.8%29%5E2%7D%5Ctimes0.2%5D%2B%5B%5Csqrt%7B1-%281%29%5E2%7D%5Ctimes0.2%5D%20%5C%5C%20%20%5C%5C%20%3D%280.9798%5Ctimes0.2%29%2B%280.9165%5Ctimes0.2%29%2B%280.8%5Ctimes0.2%29%2B%280.6%5Ctimes0.2%29%2B%280%5Ctimes0.2%29%20%5C%5C%20%20%5C%5C%20%3D0.196%2B0.183%2B0.16%2B0.12%3D0.659)
Part B:
The approximate the area of the region shown in the figure using the lower sums is given by:
![Area= [y(0)\times0.2]+[y(0.2)\times0.2]+[y(0.4)\times0.2]+[y(0.6)\times0.2] \\ +[y(0.8)\times0.2] \\ \\ =[\sqrt{1-(0)^2}\times0.2]+[\sqrt{1-(0.2)^2}\times0.2]+[\sqrt{1-(0.4)^2}\times0.2] \\ +[\sqrt{1-(0.6)^2}\times0.2] +[\sqrt{1-(0.8)^2}\times0.2] \\ \\ =(1\times0.2)+(0.9798\times0.2)+(0.9165\times0.2)+(0.8\times0.2)+(0.6\times0.2) \\ \\ =0.2+0.196+0.183+0.16+0.12=0.859](https://tex.z-dn.net/?f=Area%3D%20%5By%280%29%5Ctimes0.2%5D%2B%5By%280.2%29%5Ctimes0.2%5D%2B%5By%280.4%29%5Ctimes0.2%5D%2B%5By%280.6%29%5Ctimes0.2%5D%20%5C%5C%20%2B%5By%280.8%29%5Ctimes0.2%5D%20%5C%5C%20%5C%5C%20%3D%5B%5Csqrt%7B1-%280%29%5E2%7D%5Ctimes0.2%5D%2B%5B%5Csqrt%7B1-%280.2%29%5E2%7D%5Ctimes0.2%5D%2B%5B%5Csqrt%7B1-%280.4%29%5E2%7D%5Ctimes0.2%5D%20%5C%5C%20%2B%5B%5Csqrt%7B1-%280.6%29%5E2%7D%5Ctimes0.2%5D%20%2B%5B%5Csqrt%7B1-%280.8%29%5E2%7D%5Ctimes0.2%5D%20%5C%5C%20%5C%5C%20%3D%281%5Ctimes0.2%29%2B%280.9798%5Ctimes0.2%29%2B%280.9165%5Ctimes0.2%29%2B%280.8%5Ctimes0.2%29%2B%280.6%5Ctimes0.2%29%20%5C%5C%20%5C%5C%20%3D0.2%2B0.196%2B0.183%2B0.16%2B0.12%3D0.859)
Part C:
The approximate area of the given region is given by
Answer:
The answer is 24cm²
Step-by-step explanation:
The triangle area is given by A*½ and we get 4*2*½=6
Rectangle one is given as L*W and we solve 3*4= 12
Rectangle two is solved as L*W as well so we solved 1.5*4= 6
and we get 24cm²
