we are given two points as
A (1,5) and B (6,2)
so,
x1=1 , y1=5
x2=6 , y2=2
now, we can use distance formula
![d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=%20d%3D%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D%20%20)
now, we can plug values
and we get
![d=\sqrt{(6-1)^2+(2-5)^2}](https://tex.z-dn.net/?f=%20d%3D%5Csqrt%7B%286-1%29%5E2%2B%282-5%29%5E2%7D%20%20)
![d=\sqrt{(5)^2+(3)^2}](https://tex.z-dn.net/?f=%20d%3D%5Csqrt%7B%285%29%5E2%2B%283%29%5E2%7D%20%20)
so, we will get
.........Answer
Answer:
(-1)^n × (n!) × x^-(n+1)
Step-by-step explanation:
I've attached the solution
Answer:
He would need to deposit A.$500
Answer:
Step-by-step explanation:first you would multiply 0.88 by 300 and go all the way down the list and then add you answers