Answer:
A = 157 in^2
Step-by-step explanation:
First, we need to get the radius, we can do this by dividing our given diameter by 2.
20/2 = 10
Now, we have our radius of 10 inches.
Formula for Area of a Semi Circle: (1/2)(pi(r^2)
Input values.
(1/2)(3.14(10^2)
Solve.
(1/2)(3.14(100)
(1/2)(314)
1/2(314) = 157
Finally, we have our result which is 157 inches^2.
Answer:
x = 4, 2
Step-by-step explanation:
![(x-2)(2x-3)=(x-2)(x+1)](https://tex.z-dn.net/?f=%28x-2%29%282x-3%29%3D%28x-2%29%28x%2B1%29)
![\\\Rightarrow 2x^2-7x+6=x^2+x-2x-2\\\Rightarrow 2x^2-7x+6-x^2+x-2x-2=0\\\Rightarrow x^2-6x+8=0](https://tex.z-dn.net/?f=%5C%5C%5CRightarrow%202x%5E2-7x%2B6%3Dx%5E2%2Bx-2x-2%5C%5C%5CRightarrow%202x%5E2-7x%2B6-x%5E2%2Bx-2x-2%3D0%5C%5C%5CRightarrow%20x%5E2-6x%2B8%3D0)
While splitting the middle term the sum of the term should give us the middle term and their product should give the last term
![\\\Rightarrow x^2-4x-2x+8=0](https://tex.z-dn.net/?f=%5C%5C%5CRightarrow%20x%5E2-4x-2x%2B8%3D0)
-4-2 = -6
![-4\times -2=8](https://tex.z-dn.net/?f=-4%5Ctimes%20-2%3D8)
![\\\Rightarrow (x-4)(x-2)=0\\\Rightarrow x=4, 2](https://tex.z-dn.net/?f=%5C%5C%5CRightarrow%20%28x-4%29%28x-2%29%3D0%5C%5C%5CRightarrow%20x%3D4%2C%202)
Hence, x = 4, 2
Answer:
θ
![1.75\pi](https://tex.z-dn.net/?f=1.75%5Cpi)
r 10.551 244.151
Step-by-step explanation:
The maximum value for
is:
![\theta_{max} = \ln r](https://tex.z-dn.net/?f=%5Ctheta_%7Bmax%7D%20%3D%20%5Cln%20r)
![\theta_{max} = 2.199\pi\,rad](https://tex.z-dn.net/?f=%5Ctheta_%7Bmax%7D%20%3D%202.199%5Cpi%5C%2Crad)
The formula for the slope of the tangent line in polar coordinates is:
![m = \frac{r'\cdot \sin \theta + r \cdot \cos \theta}{r' \cdot \cos \theta - r \cdot \cos \theta}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7Br%27%5Ccdot%20%5Csin%20%5Ctheta%20%2B%20r%20%5Ccdot%20%5Ccos%20%5Ctheta%7D%7Br%27%20%5Ccdot%20%5Ccos%20%5Ctheta%20-%20r%20%5Ccdot%20%5Ccos%20%5Ctheta%7D)
Horizontal tangent lines have a slope of zero. So, the following relation must be satisfied:
![r'\cdot \sin \theta + r \cdot \cos \theta = 0](https://tex.z-dn.net/?f=r%27%5Ccdot%20%5Csin%20%5Ctheta%20%2B%20r%20%5Ccdot%20%5Ccos%20%5Ctheta%20%3D%200)
![r'\cdot \sin \theta = - r \cdot \cos \theta](https://tex.z-dn.net/?f=r%27%5Ccdot%20%5Csin%20%5Ctheta%20%3D%20-%20r%20%5Ccdot%20%5Ccos%20%5Ctheta)
![\tan \theta = - \frac{r}{r'}](https://tex.z-dn.net/?f=%5Ctan%20%5Ctheta%20%3D%20-%20%5Cfrac%7Br%7D%7Br%27%7D)
![\tan \theta = -\frac{e^{\theta}}{e^{\theta}}](https://tex.z-dn.net/?f=%5Ctan%20%5Ctheta%20%3D%20-%5Cfrac%7Be%5E%7B%5Ctheta%7D%7D%7Be%5E%7B%5Ctheta%7D%7D)
![\tan \theta = -1](https://tex.z-dn.net/?f=%5Ctan%20%5Ctheta%20%3D%20-1)
![\theta = \tan^{-1}(-1)](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Ctan%5E%7B-1%7D%28-1%29)
, for all
.
The maximum value of i is:
![i = \frac{\theta_{max}-\frac{3}{4}\pi }{\pi}](https://tex.z-dn.net/?f=i%20%3D%20%5Cfrac%7B%5Ctheta_%7Bmax%7D-%5Cfrac%7B3%7D%7B4%7D%5Cpi%20%7D%7B%5Cpi%7D)
![i = \frac{2.199-0.75}{1}](https://tex.z-dn.net/?f=i%20%3D%20%5Cfrac%7B2.199-0.75%7D%7B1%7D)
(
).
Then, values are listed below:
θ
![1.75\pi](https://tex.z-dn.net/?f=1.75%5Cpi)
r 10.551 244.151
Answer: c) SAS Postulate
<u>Step-by-step explanation:</u>
DP = PC Sides are congruent
∠ADP ≡ ∠BCP Angles are congruent (angles are <u>between</u> the sides)
AD = BC Sides are congruent
To finish the proof, we can state that ΔADP ≡ ΔBCP by the Side-Angle-Side (SAS) Postulate
Answer:
See below.
Step-by-step explanation:
7.)
Supplementary angles add up to 180°. ∠B's supplement is 22°. This means ∠B is 158°. Since ∠A ≅ ∠B, the m∠A = 158°.
8.)
Again, supplementary angles add up to 180°. ∠P is a right angle meaning it is 90°. Therefore, the m∠Q is 90° as well.
9.)
Complementary angles add up to 90°. In order for ∠S and ∠T to be complementary, they both need to angle measurements less than 90° meaning they both need to be acute angles.
Now that we know that ∠T is acute, the only way ∠U can be its supplement is if it has an angle measure greater than 90°. Therefore, ∠U is an obtuse angle.
10.)
Similar to question 9, if an angle is obtuse, its supplement must be an acute angle.