Hey op, what does this m e a n
Answer:a place that the government controls
Step-by-step explanation:
Points A, B, C, D, and O have coordinates (0, 3), (4, 4), (4, y), (4, 0), and (0, 0), respectively. The area of ACDO is 5 times
belka [17]
Solution:
Area of Quadrilateral ACDO= Area (ΔAOD)+Area(ΔACD)-----(1)
Area of Quadrilateral ABDO= Area (ΔAOD)+Area(ΔABD)------(2)
⇒ (1) = 5 ×(2)→→→→GIven
→Area (ΔAOD)+Area(ΔACD)=5 Area (ΔAOD)+5 Area(ΔABD),
→5 Area (ΔABD)+4 Area (ΔAOD)-Area (ΔACD)=0-----(3)
As, Area of a Triangle ![={\text{having vertices}} (x_{1},y_{1}),(x_{2},y_{2}) {\text{and}} (x_{3},y_{3})=\frac{1}{2}\times[x_{1}(y_{2}-y_{3})-x_2(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]](https://tex.z-dn.net/?f=%3D%7B%5Ctext%7Bhaving%20vertices%7D%7D%20%28x_%7B1%7D%2Cy_%7B1%7D%29%2C%28x_%7B2%7D%2Cy_%7B2%7D%29%20%7B%5Ctext%7Band%7D%7D%20%28x_%7B3%7D%2Cy_%7B3%7D%29%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%5Bx_%7B1%7D%28y_%7B2%7D-y_%7B3%7D%29-x_2%28y_%7B3%7D-y_%7B1%7D%29%2Bx_%7B3%7D%28y_%7B1%7D-y_%7B2%7D%29%5D)
Area (ΔACD)
![=\frac{1}{2}[0(y-0)-4(0-3)+4(3-y)]\\\\= \frac{1}{2}[12+12-4y]\\\\=12-2 y](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%5B0%28y-0%29-4%280-3%29%2B4%283-y%29%5D%5C%5C%5C%5C%3D%20%5Cfrac%7B1%7D%7B2%7D%5B12%2B12-4y%5D%5C%5C%5C%5C%3D12-2%20y)
Area (ΔABD)
![=\frac{1}{2}[0(4-0)-4(0-3)+4(3-4)]\\\\ =\frac{1}{2}[12-4]=4](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%5B0%284-0%29-4%280-3%29%2B4%283-4%29%5D%5C%5C%5C%5C%20%3D%5Cfrac%7B1%7D%7B2%7D%5B12-4%5D%3D4)
Area (ΔAOD)![=\frac{1}{2}[0(3-0)-0(0-0)+4(0-3)]=6](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%5B0%283-0%29-0%280-0%29%2B4%280-3%29%5D%3D6)
Putting these values in equation (3)
→20+24-12+2 y=0
→ 2 y= -32
Dividing both sides by , 2 we get
→y= -16
Step-by-step explanation:
yes, if your teacher meant to include the default factors of every positive integer number : 1 and the number itself.
these are the 2 positive integer factors.
if a positive number can only be divided without remainder by 1 or by itself (with result 1), then it is per definition a prime number.
if the number can be divided without remainder by another number, then we are not dealing with a prime number.
but if your teacher did not have these default factors in mind, then no :
example : 15
15 has only 2 factors : 3 and 5.
both are positive integer numbers.
and yet 15 is not a prime number.
but back to my first point, 15 also has formally the factors 1 and 15. so, if we count them too, then we have suddenly 4 factors. and the original statement fits again.
