Answer:
MP = 17
Step-by-step explanation:
Since N is the midpoint of MO then MN = NO = 7
NO + OP = NP , then
7 + OP = 10 ( subtract 7 from both sides )
OP = 3
Then
MP = MN + NO + OP = 7 + 7 + 3 = 17
Answer:
The measures of each would be:
105° and 75°
Step-by-step explanation:
Supplementary angles are two angles whose measures sum up to 180° or they form a straight line.
So if an angle measures 30° less than the measure of its supplementary, it wold mean that both angles together is equal to 180°.
∠1 = x
∠2 = x-30°
∠1 + ∠2 = 180°
So here we plug in our equations:
∠1 + ∠2 = 180°
x + x - 30° = 180°
2x - 30° = 180°
We solve for the x then:
Add 30° on both sides of the equation:
2x - 30° + 30° = 180° + 30°
2x = 210°
Divide both sides by 2:
2x/2 = 210°/2
x = 105°
∠1 = 105°
Now we solve for the second angle:
∠1 + ∠2 = 180°
105° + ∠2 = 180°
Subtract 105° from both sides of the equation:
105° + ∠2 - 105° = 180° - 105°
∠2 = 75°
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
it will be 63.1(56TAN) or 93.549