This is a function y=√x that was moved 2 units to the right.
y=√(x-2)
4(x + 3)² = 25(x - 2)²
Square root both sides:
2(x+ 3) = 5(x - 2)
Distribute 2 and 5:
2x + 6 = 5x - 10
Take away 2x from both sides:
6 = 3x - 10
Add 10 to both sides:
16 = 3x
Switch sides:
3x = 6
Divide both sides by 3:
x = 2
Answer: x = 2
![\bf \textit{we know the range for }cos\left( \frac{1}{x} \right)\textit{ is }[-1,1]\textit{ therefore} \\\\\\ -1~\ \textless \ ~cos\left( \frac{1}{x} \right)~\ \textless \ ~1\impliedby \textit{multiplying all sides by }x^2 \\\\\\ -1x^2~\ \textless \ ~x^2cos\left( \frac{1}{x} \right)~\ \textless \ ~1x^2\implies -x^2~\ \textless \ ~x^2cos\left( \frac{1}{x} \right)~\ \textless \ ~x^2](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bwe%20know%20the%20range%20for%20%7Dcos%5Cleft%28%20%5Cfrac%7B1%7D%7Bx%7D%20%5Cright%29%5Ctextit%7B%20is%20%7D%5B-1%2C1%5D%5Ctextit%7B%20therefore%7D%0A%5C%5C%5C%5C%5C%5C%0A-1~%5C%20%5Ctextless%20%5C%20~cos%5Cleft%28%20%5Cfrac%7B1%7D%7Bx%7D%20%5Cright%29~%5C%20%5Ctextless%20%5C%20~1%5Cimpliedby%20%5Ctextit%7Bmultiplying%20all%20sides%20by%20%7Dx%5E2%0A%5C%5C%5C%5C%5C%5C%0A-1x%5E2~%5C%20%5Ctextless%20%5C%20~x%5E2cos%5Cleft%28%20%5Cfrac%7B1%7D%7Bx%7D%20%5Cright%29~%5C%20%5Ctextless%20%5C%20~1x%5E2%5Cimplies%20-x%5E2~%5C%20%5Ctextless%20%5C%20~x%5E2cos%5Cleft%28%20%5Cfrac%7B1%7D%7Bx%7D%20%5Cright%29~%5C%20%5Ctextless%20%5C%20~x%5E2)
if the limit of -x² goes to "something", and the limit of x² goes to the same "something", if their limit coincide, and yet they're bounding the cosine expression, therefore, since the cosine expression is "sandwiched" between -x² and x², then the cosine expression "squeezes in" that little sliver between both -x² and x², and will inevitably go to the same limit.
Answer:
B. 3/5
Step-by-step explanation:
Answer:
what's the rest of the problem?