Answer: ![\sqrt[5]{y}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7By%7D)
I realize its probably not the largest readable font. If you are having trouble reading it, it is the square root of y; however, there is a tiny little 5 in the upper left corner to indicate a fifth root. So you would read it out as "the fifth root of y"
The rule I'm using is
![x^{1/n} = \sqrt[n]{x}](https://tex.z-dn.net/?f=x%5E%7B1%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%7D)
and the more general rule we could use is
![x^{m/n} = \sqrt[n]{x^m}](https://tex.z-dn.net/?f=x%5E%7Bm%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D)
where m = 1. This rule helps convert from rational exponent form (aka fractional exponents) to radical form.
<h3>Answer: Choice D</h3>
Divide both sides of the first equation by 7, then add the result to the second equation
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Explanation:
We can multiply both sides of an equation by the same number and the equation will be equivalent to the original.
For example, if we had x = 5, then we could get 2x = 10 after multiplying both sides by 2. The reason they are equivalent equations is because the same x value is the solution for both equations.
We can also divide both sides of an equation by the same number and the two equations would be equivalent. We can go from 2x = 10 back to x = 5 when we divide both sides by 2.
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If we divide both sides of 7x - 21y = 14 by 7, then we end up with x - 3y = 2. Simply divide each term (7x, -21y, and 14) by 7.
Because 7x-21y=14 and x-3y=2 are equivalent, this means we can replace the "7x-21y=14" with "x-3y=2"
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The new system would be
x-3y = 2
2x+3y = 11
From here you add straight down. Doing so will have the y terms add to -3y+3y = 0y = 0. After this point the y terms are eliminated and you can solve for x just like with any other equation of one variable.
How you do this is you go from left to right with each one, and figure out which number is greater. For this particular problem the answer will be,
1.2 1.23 2.31 3.2
Answer: 105
Step-by-step explanation: 25 frames is 1 second. The video already played 3 and 2/5 seconds before the player started to count 0, so you can write 3 and 2/5 seconds as an improper fraction. =(5*3)+2 / 5 = 17/5 seconds. Multiply by 25 frame. 17/5 *25 =85 frames. So according to the video counter,after 17/5 seconds,it should count 85 frames.However,at 0 seconds,it indicated a count of 190 frames. then to get the number of frames that were already in count you would subtract 85 frames from the 190 frames.
190-85=105 frames
(Not 100% sure so apologies if I'm wrong)
