The correct answer is the letter B
Palelogram area:
Knowing the base and the height, the area of the parallelogram is calculated by its formula:
A1 = b * h
Where
b: base
h: height.
We have then
b = root ((- 4 - (- 10)) ^ 2+ (2-2) ^ 2) = 6
h = root ((- 7 - (- 7)) ^ 2+ (3-2) ^ 2) = 1
A1 = (6) * (1) = 6 units ^ 2
Rectangle area
A2 = L * W
Where
L: long
W: Wide
We have then:
L = root ((1 - (- 2)) ^ 2 + (- 3 - (- 4)) ^ 2) = 3.16227766
W = root ((- 4 - (- 2)) ^ 2+ (2 - (- 4)) ^ 2) = 6.32455532
A2 = (3.16227766) * (6.32455532)
A2 = 20 units ^ 2
The total area is the sum of both ares:
A = A1 + A2
A = 6 + 20
A = 26 units ^ 2
Answer:
the area of this figure
A = 26 units ^ 2
Step-by-step explanation:
Statement:
2-) ∠BAC = ∠EDC
<em>Reason:</em>
Angles opposite to equal sides of a triangle are equal (Alternate Interior Angles Theorem)
Statement:
3-) AC = CD
<em>Reason:</em>
CPCTC ("Corresponding Parts of Congruent Triangles are Congruent")
Statement:
4-) ∠BCA = ∠DCE
<em>Reason:</em>
Vertical Angles Theorem (states that vertical angles, angles that are opposite each other and formed by two intersecting straight lines, are congruent)
Statement:
5-) triangle ABC = triangle DEC
ASA Postulate
The ASA (Angle-Side-Angle) postulate states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent. (The included side is the side between the vertices of the two angles.)
<h2>22</h2><h3>Answer: B</h3><h3 /><h2>23</h2><h3>Answer: D</h3><h3 /><h2>24</h2><h3>Answer: A</h3><h3 /><h2>25</h2><h3>Answer: C</h3>
![\sqrt{x-8}](https://tex.z-dn.net/?f=%5Csqrt%7Bx-8%7D)
is undefined if the argument
![x-8](https://tex.z-dn.net/?f=x-8)
is negative, so you first need to require that
![x-8\ge0\implies x\ge8](https://tex.z-dn.net/?f=x-8%5Cge0%5Cimplies%20x%5Cge8)
We're not done yet, though, because
![\dfrac1{\sqrt{x-8}}](https://tex.z-dn.net/?f=%5Cdfrac1%7B%5Csqrt%7Bx-8%7D%7D)
still doesn't exist when
![x=8](https://tex.z-dn.net/?f=x%3D8)
, so we remove this from the domain and we're left with
![x>8](https://tex.z-dn.net/?f=x%3E8)
, or in interval notation,
![(8,\infty)](https://tex.z-dn.net/?f=%288%2C%5Cinfty%29)
To find the range, consider the limits of the function as you approach either endpoint of the domain.
![\displaystyle\lim_{x\to8^+}\frac1{\sqrt{x-8}}=+\infty](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto8%5E%2B%7D%5Cfrac1%7B%5Csqrt%7Bx-8%7D%7D%3D%2B%5Cinfty)
![\displaystyle\lim_{x\to\infty}\frac1{\sqrt{x-8}}=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto%5Cinfty%7D%5Cfrac1%7B%5Csqrt%7Bx-8%7D%7D%3D0)
Since
![\dfrac1{\sqrt{x-8}}](https://tex.z-dn.net/?f=%5Cdfrac1%7B%5Csqrt%7Bx-8%7D%7D)
is positive everywhere, the range is