<h3>
Answer: shaded region has area 107 square feet</h3>
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Work Shown:
A = area of larger rectangle
A = length*width = 9*15 = 135 square feet
B = area of smaller rectangle
B = length*width = 7*4 = 28 square feet
C = area of shaded region
C = difference in areas of A and B
C = A - B = 135 - 28 = 107 square feet
The 2 like terms are 400km +924m
Given:
A(-5,4)
B(3,4)
C(3,-5)
So point D is:
so point D is (-5,-5)
For AB is
Distance between two point is:
![\begin{gathered} (x_1,y_1)and(x_2,y_2) \\ D=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%28x_1%2Cy_1%29and%28x_2%2Cy_2%29%20%5C%5C%20D%3D%5Csqrt%5B%5D%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D%20%5Cend%7Bgathered%7D)
so distance between A(-5,4) and B(3,4) is:
![\begin{gathered} D=\sqrt[]{(3-(-5))^2+(4-4)^2} \\ =\sqrt[]{(8)^2+0^2} \\ =8 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20D%3D%5Csqrt%5B%5D%7B%283-%28-5%29%29%5E2%2B%284-4%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B%288%29%5E2%2B0%5E2%7D%20%5C%5C%20%3D8%20%5Cend%7Bgathered%7D)
So AB is 8 unit apart.
For B(3,4) and C(3,-5).
![\begin{gathered} D=\sqrt[]{(3-3)^2+(-5-4)^2} \\ =\sqrt[]{0^2+(-9)^2} \\ =9 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20D%3D%5Csqrt%5B%5D%7B%283-3%29%5E2%2B%28-5-4%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B0%5E2%2B%28-9%29%5E2%7D%20%5C%5C%20%3D9%20%5Cend%7Bgathered%7D)
So BC is 9 unit apart.
For fourth bush point is (-5,-5) it left of point C(3,-5) is:
![\begin{gathered} D=\sqrt[]{(3-(-5))^2+(-5-(-5))^2} \\ =\sqrt[]{(8)^2+0^2} \\ =8 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20D%3D%5Csqrt%5B%5D%7B%283-%28-5%29%29%5E2%2B%28-5-%28-5%29%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B%288%29%5E2%2B0%5E2%7D%20%5C%5C%20%3D8%20%5Cend%7Bgathered%7D)
so fourth bush is 8 unit left of C.
For fourth bush(-5,-5) below to point A(-5,4)
![\begin{gathered} D=\sqrt[]{(-5-(-5))^2+(4-(-5))^2} \\ =\sqrt[]{0^2+9^2} \\ =9 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20D%3D%5Csqrt%5B%5D%7B%28-5-%28-5%29%29%5E2%2B%284-%28-5%29%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B0%5E2%2B9%5E2%7D%20%5C%5C%20%3D9%20%5Cend%7Bgathered%7D)
so fourth bush 9 units below of A.
Answer:
It's B
Step-by-step explanation:
2 times 5 = 10
0.8 times 5=4
10 + 4 =14
14-3=11
Answer:
1. D
2. B
3. A
Step-by-step explanation:
Question 1:
The pair of <JKL and <LKM can be referred to as linear pairs. They are two adjacent angles that are formed from the intersecting of two lines.
Question 2:
Given that <KLM = x°
<KML = 50°
<JKL = (2x - 15)°
According to the exterior angle theorem, exterior ∠ JKL = <KLM + KML.
2x - 15 = x + 50
Solve for x
2x - x = 15 + 50
x = 65
Therefore, <KLM = 65°
QUESTION 3:
<JKL = 2x - 15
Plug in the value of x
<JKL = 2(65) - 15
= 130 - 15
<JKL = 115°
