So you know the new equation must have the same slope as the old equation since they're parallel to one another. Next just simply take the slope and the point given and plug them into the point-slope formula and you have your new equation:
y-y1=m(x-x1) --- Point-Slope Formula
y+5=2(x-4)
y+5=2x-8
y=2x-13 --- Answer
If the speed is constant, there is no acceleration. No acceleration, no force.
Answer:
$969.04
Step-by-step explanation:
now at age of 18 Brad has 1500 * 1,065^18
at age of 21 he will have 1500 * 1,065^21
if he waits the difference will be 1500 * 1,065^21 - 1500 * 1,065^18 = $969.04
The sea level is considered to have an elevation of zero feet.
To determine which checkpoint is furthest away from the sea level you have to calculate their absolute difference to zero and compare them:
Checkpoint A
![\begin{gathered} d_A=|15.6-0| \\ d_A=15.6ft \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20d_A%3D%7C15.6-0%7C%20%5C%5C%20d_A%3D15.6ft%20%5Cend%7Bgathered%7D)
Checkpoint B
![\begin{gathered} d_B=|17.1-0| \\ d_B=17.1ft \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20d_B%3D%7C17.1-0%7C%20%5C%5C%20d_B%3D17.1ft%20%5Cend%7Bgathered%7D)
Checkpoint C
![\begin{gathered} d_C=|5.2-0| \\ d_C=5.2ft \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20d_C%3D%7C5.2-0%7C%20%5C%5C%20d_C%3D5.2ft%20%5Cend%7Bgathered%7D)
Checkpoint D
![\begin{gathered} d_D=|-6.5-0| \\ d_D=6.5ft \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20d_D%3D%7C-6.5-0%7C%20%5C%5C%20d_D%3D6.5ft%20%5Cend%7Bgathered%7D)
Checkpoint E
![\begin{gathered} d_E=|-18.5-0| \\ d_E=18.5ft \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20d_E%3D%7C-18.5-0%7C%20%5C%5C%20d_E%3D18.5ft%20%5Cend%7Bgathered%7D)
Ordered from closest to furthest the distance of each checkpoint from sea level is:
[tex]0The furthest checkpoint to sea level is Checkpoint E, which is 18.5ft away