2016年4月17日 · sqrt((5+x)(5-x)) sqrt(25-x^2)=sqrt((5+x)(5-x))

2018年4月10日 · Solution: -5 <= x <= 5 or [-5,5] 25-x^2 >= 0 or (5+x)(5-x)>=0 Critical points are 5+x=0 or x= -5 and 5-x=0 or x= 5 f(x)=0 when x=-5 and x=5 Sign chart: When x< -5 ...

2015年4月15日 · You must apply x=5/costheta and do the mathematics step by step. This integral does not seem solvable by other techniques, e.g. integration by part, just integration by …

2016年1月21日 · In the given expression #x^2+25# we note that the discriminant #sqrt(b^2-4ac)# is a negative quantity. As such the quadratic has only imaginary roots. These can be …

2015年7月29日 · Notice how sqrt(25 + x^2) prop sqrt(a^2 + x^2), which implies x = atantheta with a = 5. If we let: x = 5tantheta dx = 5sec^2thetad theta sqrt(25 + x^2) = sqrt(5^2 + …

2020年7月23日 · This is where I found the discepancy. The online solution has the basic equation as $$1=A*x(x^2+25)+B*(x^2 ...

2016年4月15日 · As xto5-, sqrt(25-x^2)to0. Unless otherwise stated, the graph of y = sqrt(25-x^2) means y=+-sqrt(25-x^2) and represents the circle with center at the origin and radius = 5. If it …

2016年11月1日 · How do you evaluate the limit #(x^2-25)/(x^2-4x-5)# as x approaches 5? Calculus Limits Determining Limits Algebraically. 1 Answer

2017年8月4日 · - (x + 5)/(x - 5) (25 - x^2)/(x^2 - 10x + 25) = ((5 - x)(5 + x))/((x - 5)(x-5)) =(color(blue)(-cancel((x-5)))(5 + x))/(cancel((x - 5))(x-5)) = - (x + 5)/(x - 5)

2020年12月9日 · I thought your $(25-x^2)^{3/2}$ was in the denominator, which is why I got a different value from you. I've edited it to reflect your work. I checked my final answer, and it's …