For number one the second card must be positive or he would have to put it back if its negative for number 2 the card must be 3 4 5 because negative 2 plus 3 is 1 which is positive and negative 2 plus 4 and 5 would allow her to keep her cards. for the third one danny could have pulled a negaitive 1 2 3 4 and 5. oh dude i forgot to put decimals but still the same thing
See, to make a 5-letter word , 5 places are to be filled ___ ___ ___ ___ ___ 1st 2nd 3rd 4th 5th the first place can be filled in 5 ways (s,p,e,l,l) 2nd place in 4 ways. 3rd place in 3 ways. 4th place in 2 ways. 5th place in 1 way. now , ther are 2L's. hence total no. of words that can be formed=(5*4*3*2*1) / 2*1
Answer:
6x=48
7y-1=48 (im assuming the y is on the 7.)
48/6=8=x
48-1=47/7=6.7=y
Answer:
Choice B
Step-by-step explanation:
Given radical expression:
![\sqrt[4]{1296 {x}^{16} {y}^{12} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B1296%20%7Bx%7D%5E%7B16%7D%20%20%7By%7D%5E%7B12%7D%20%7D%20)
To Find:
The Simpler form of this expression
Soln:
![= \sqrt[4]{1296 {x}^{16} {y}^{12} }](https://tex.z-dn.net/?f=%20%3D%20%20%5Csqrt%5B4%5D%7B1296%20%7Bx%7D%5E%7B16%7D%20%20%7By%7D%5E%7B12%7D%20%7D%20)
We could re-write the given expression, according to the law of exponents:
![= \tt \sqrt[4]{(6x {}^{4}y {}^{3}) {}^{4} }](https://tex.z-dn.net/?f=%20%3D%20%20%5Ctt%20%5Csqrt%5B4%5D%7B%286x%20%7B%7D%5E%7B4%7Dy%20%7B%7D%5E%7B3%7D%29%20%20%7B%7D%5E%7B4%7D%20%20%7D%20)
Now we need to bring terms out of the radical as:
Bring out 6x^4 from the absolute & put y^3 only in it:
Choice B is accurate.
Here is my answer, hope to help you
