Plz help me with this!!!

one light flashes every 2 minutes and another flashes every 7 minutes. If both lights flash together at 1pm, what is the first t

vovangra [49]

Light A flashes every 2 minutes while light B flashes every 7. The goal is to figure out when both lights flash at the same time and then figure out the other part later. You have to find the common multiples of 7 and 2. Meaning that you have to find out which number can they both be multiplied into. In this case, it is 14 because 7 can be multiplied by 2 to get 14 and 2 can be multiplied by 7. So if every 14 minutes they flash together at the same time, now you need to find out what time AFTER 3 will they both flash. You know that they both flashed at 1:00 so they will flash again at 1:14 and 1:28 and so on, but you need to find out when is the soonest they will flash after 3. 2 hours after the 1:00 flash will be 3:00 so that is 120 minutes. 120 minutes divided by 14 minutes comes to 8.5 but you dont need the decimal. Just simply multiply 14 by 8 and that comes to 112 minutes and if you add that to 1:00 you get 2:52. so they flashed at 2:52 so the next time they flash will be 14 minutes after 2:52. Sorry for the super long answer

6 0

3 years ago

Simplify sin^2y/sec^2 y−1 to a single trigonometric function

liubo4ka [24]

Answer:

$\frac{ { \sin}^{2} y}{ { \sec}^{2}y - 1 } = { \cos }^{2} y$

Step-by-step explanation:

We know that ${ \tan }^{2} y = { \sec }^{2} y - 1$

Also , ${ \tan}^{2} y = \frac{ { \sin }^{2} y}{ { \cos }^{2}y }$

So ,

$\frac{ { \sin }^{2}y }{ { \sec }^{2}y - 1 } = \frac{ { \sin}^{2} y}{ { \tan }^{2} y} = \frac{ { \sin }^{2}y }{ \frac{ { \sin}^{2} y}{ { \cos}^{2}y } } = { \cos }^{2} y$