The first step is to determine the distance between the points, (1,1) and (7,9)
We would find this distance by applying the formula shown below
![\begin{gathered} \text{Distance = }\sqrt[]{(x2-x1)^2+(y2-y1)^2} \\ \text{From the graph, } \\ x1\text{ = 1, y1 = 1} \\ x2\text{ = 7, y2 = 9} \\ \text{Distance = }\sqrt[]{(7-1)^2+(9-1)^2} \\ \text{Distance = }\sqrt[]{6^2+8^2}\text{ = }\sqrt[]{100} \\ \text{Distance = 10} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctext%7BDistance%20%3D%20%7D%5Csqrt%5B%5D%7B%28x2-x1%29%5E2%2B%28y2-y1%29%5E2%7D%20%5C%5C%20%5Ctext%7BFrom%20the%20graph%2C%20%7D%20%5C%5C%20x1%5Ctext%7B%20%3D%201%2C%20y1%20%3D%201%7D%20%5C%5C%20x2%5Ctext%7B%20%3D%207%2C%20y2%20%3D%209%7D%20%5C%5C%20%5Ctext%7BDistance%20%3D%20%7D%5Csqrt%5B%5D%7B%287-1%29%5E2%2B%289-1%29%5E2%7D%20%5C%5C%20%5Ctext%7BDistance%20%3D%20%7D%5Csqrt%5B%5D%7B6%5E2%2B8%5E2%7D%5Ctext%7B%20%3D%20%7D%5Csqrt%5B%5D%7B100%7D%20%5C%5C%20%5Ctext%7BDistance%20%3D%2010%7D%20%5Cend%7Bgathered%7D)
Distance = 10 units
If one unit is 70 meters, then the distance between both entrances is
70 * 10 = 700 meters
Paralell has same slope
y=mx+b
m=slope
y=5x-1
when is the slope 5?
take the derivitive of 2e^x
f'(2e^x)=2f'(e^x)=2e^x
that is the slope
5=2e^x
divide both sides by 2
2.5=e^x
take the ln of both sides
ln2.5=x
when x=ln2.5 they are paralell
The 20th visitor will be the first to get both the key chain and the bumper sticker, I found this out by finding the least common multiple, or LCM, you do this by writing out the multiples of both 4 and 10, 4 8 12 16 20..... and 10 20 30... then you find the smallest number that both 4 and 10 have in common. I hope this helps, sorry if it doesn't.
Answer:
The answer would be 50
Step-by-step explanation:
Step-by-step explanation:
verticales están a 12 metros uno del otro. Si tanto la parte superior como la inferior del arco son arcos de parábola, redondeo hasta el metro más cercano la suma de longitudes de los tirantes verticales e inclinados.
