Answer:
hd wallpapers for mobile under the one year old girl in india and the world of the world is known as the world of india is the first part and the part of a girl
Answer:
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Step-by-step explanation:
Answer:
y-coordinate is 5 or -1.
Step-by-step explanation:
Point A is at (x, 2) and B is at (x+6, 2). Since AB must lie on the line y=2 and be 6 units long. Point C is on the line x = -3 . So let C be at (-3, y).
Since ΔABC is a right angle, then point C must have the same x-coordinate as point A. Therefore, A(-3, 2) and B(2, 2).
The area of ΔABC is 6. So,
9 = 1/2 (b)(h)
where b is the base and h is the height.
so b = 6 and h = AC
Solving this for C gives
9 = 1/2 (6)(AC)
18/6 = AC
3 = AC
9 = 1/2 (6)(AC)
18/6 = AC
3 = AC
Point C must lie 3 units above point A or 3 units below the point A. If it lies 3 units above, then it has a y-coordinate of 2 + 3 = 5.
If it lies 3 units below, it has a y-coordinate 2 - 3 = -1.
Therefore, y-coordinate is 5 or -1.
Step-by-step explanation:
Answer:
![\dfrac{1}{2x(x-1)}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2x%28x-1%29%7D)
Step-by-step explanation:
Given
![\dfrac{x^2+2x+1}{x^2-1}\div (2x^2+2x)](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E2%2B2x%2B1%7D%7Bx%5E2-1%7D%5Cdiv%20%282x%5E2%2B2x%29)
Consider the numerator:
![x^2+2x+1=(x+1)^2](https://tex.z-dn.net/?f=x%5E2%2B2x%2B1%3D%28x%2B1%29%5E2)
Consider the denominator:
![x^2-1=(x-1)(x+1)](https://tex.z-dn.net/?f=x%5E2-1%3D%28x-1%29%28x%2B1%29)
Hence, the fraction becomes
![\dfrac{(x+1)^2}{(x-1)(x+1)}=\dfrac{x+1}{x-1}](https://tex.z-dn.net/?f=%5Cdfrac%7B%28x%2B1%29%5E2%7D%7B%28x-1%29%28x%2B1%29%7D%3D%5Cdfrac%7Bx%2B1%7D%7Bx-1%7D)
Consider the expression in brackets:
![2x^2+2x=2x(x+1)](https://tex.z-dn.net/?f=2x%5E2%2B2x%3D2x%28x%2B1%29)
Divide:
![\dfrac{x^2+2x+1}{x^2-1}\div (2x^2+2x)=\dfrac{x+1}{x-1}\div 2x(x+1)=\dfrac{x+1}{x-1}\times \dfrac{1}{2x(x+1)}=\dfrac{1}{2x(x-1)}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E2%2B2x%2B1%7D%7Bx%5E2-1%7D%5Cdiv%20%282x%5E2%2B2x%29%3D%5Cdfrac%7Bx%2B1%7D%7Bx-1%7D%5Cdiv%202x%28x%2B1%29%3D%5Cdfrac%7Bx%2B1%7D%7Bx-1%7D%5Ctimes%20%5Cdfrac%7B1%7D%7B2x%28x%2B1%29%7D%3D%5Cdfrac%7B1%7D%7B2x%28x-1%29%7D)
Given:
Vertices of a square are A(-4,6), B(5,6) C(4,-2), and D(-5,-2).
To find:
The intersection of the diagonals of square ABCD.
Solution:
We know that diagonals of a square always bisect each other. It means intersection of the diagonals of square is the midpoint of diagonals.
In the square ABCD, AC and BD are two diagonals. So, intersection of the diagonals is the midpoint of both AC and BD.
We can find midpoint of either AC or BD because both will result the same.
Midpoint of A(-4,6) and C(4,-2) is
![Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)](https://tex.z-dn.net/?f=Midpoint%3D%5Cleft%28%5Cdfrac%7Bx_1%2Bx_2%7D%7B2%7D%2C%5Cdfrac%7By_1%2By_2%7D%7B2%7D%5Cright%29)
![Midpoint=\left(\dfrac{-4+4}{2},\dfrac{6+(-2)}{2}\right)](https://tex.z-dn.net/?f=Midpoint%3D%5Cleft%28%5Cdfrac%7B-4%2B4%7D%7B2%7D%2C%5Cdfrac%7B6%2B%28-2%29%7D%7B2%7D%5Cright%29)
![Midpoint=\left(\dfrac{0}{2},\dfrac{6-2}{2}\right)](https://tex.z-dn.net/?f=Midpoint%3D%5Cleft%28%5Cdfrac%7B0%7D%7B2%7D%2C%5Cdfrac%7B6-2%7D%7B2%7D%5Cright%29)
![Midpoint=\left(\dfrac{0}{2},\dfrac{4}{2}\right)](https://tex.z-dn.net/?f=Midpoint%3D%5Cleft%28%5Cdfrac%7B0%7D%7B2%7D%2C%5Cdfrac%7B4%7D%7B2%7D%5Cright%29)
![Midpoint=\left(0,2\right)](https://tex.z-dn.net/?f=Midpoint%3D%5Cleft%280%2C2%5Cright%29)
Therefore, the intersection of the diagonals of square ABCD is (0,2).