at the beginning of year 4, only 3 years have elapsed, the 4th year hasn't started yet, since it's at the beginning, so at the beginning of year 4 we can say only 4-1 years have elapsed.
![A=700\left( 1 + \frac{0.05}{1} \right)^{1\cdot 3}\implies A = 700(1+0.05)^3\implies A(4)=700(1+0.05)^{4-1} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill A(n)=700(1+0.05)^{n-1}~\hfill](https://tex.z-dn.net/?f=A%3D700%5Cleft%28%201%20%2B%20%5Cfrac%7B0.05%7D%7B1%7D%20%5Cright%29%5E%7B1%5Ccdot%203%7D%5Cimplies%20A%20%3D%20700%281%2B0.05%29%5E3%5Cimplies%20A%284%29%3D700%281%2B0.05%29%5E%7B4-1%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20A%28n%29%3D700%281%2B0.05%29%5E%7Bn-1%7D~%5Chfill)
Answer:
15-3r
Step-by-step explanation:
A expression cannot have a equal sign only = + or -
If you have questions feel free to ask me! :)
Answer:
Perpendicular lines are lines that intersect at a right (90 degrees) angle.so no form of perpendicular for angles except 90° angles
Step-by-step explanation:
Two lines are perpendicular if and only if they form a right angle. Perpendicular lines (or segments) actually form four right angles, even if only one of the right angles is marked with a box.
Use Pythagorean theorem (a squared is equal to c squared minus b squared)
Where a is the unknown leg, c is the hypotenuse and b is the known leg
a^2 = 13^2-5^2 which is 144
Then square root one hundred and fourty four to get rid of the power on a
Square root of one hundred and fourty four is twelve
