Question
Find the least common multiple of 8 and 15.
Answer= 120
Answer:
Perimeters = 3 in : 7 in
Areas = 9 in² : 49 in²
Step-by-step explanation:
<em>If the figures are </em><em>squares</em><em>, and the measurements are the length of that sides (as you have said), then we will do the following. </em>
[The left square]
P = 4a
P = 4(15)
P = 60 in
-
A = a²
A = 15²
A = 225 in²
[The right square]
P = 4a
P = 4(35)
P = 140 in
-
A = a²
A = (35)²
A = 1,225 in²
[Simplifying ratios]
Perimeters = 60 in : 140 in
Perimeters = 6 in : 14 in
Perimeters = 3 in : 7 in
-
Areas = 225 in² : 1,225 in²
Areas = 45 in² : 245 in²
Areas = 9 in² : 49 in²
The position function of a particle is given by:

The velocity function is the derivative of the position:

The particle will be at rest when the velocity is 0, thus we solve the equation:

The coefficients of this equation are: a = 2, b = -9, c = -18
Solve by using the formula:
![t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D)
Substituting:
![\begin{gathered} t=\frac{9\pm\sqrt[]{81-4(2)(-18)}}{2(2)} \\ t=\frac{9\pm\sqrt[]{81+144}}{4} \\ t=\frac{9\pm\sqrt[]{225}}{4} \\ t=\frac{9\pm15}{4} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20t%3D%5Cfrac%7B9%5Cpm%5Csqrt%5B%5D%7B81-4%282%29%28-18%29%7D%7D%7B2%282%29%7D%20%5C%5C%20t%3D%5Cfrac%7B9%5Cpm%5Csqrt%5B%5D%7B81%2B144%7D%7D%7B4%7D%20%5C%5C%20t%3D%5Cfrac%7B9%5Cpm%5Csqrt%5B%5D%7B225%7D%7D%7B4%7D%20%5C%5C%20t%3D%5Cfrac%7B9%5Cpm15%7D%7B4%7D%20%5Cend%7Bgathered%7D)
We have two possible answers:

We only accept the positive answer because the time cannot be negative.
Now calculate the position for t = 6:
Hope this helps!!! It's how I found it I hope it's right