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 trigonometric substitution. On integration by trigonometric substitution, you either apply trigonometry or trigonometric functions, such as the one herein. Unfortunately there is no …

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 calculated and factors found as

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 + 5^2tan^2theta) = 5sectheta Then we get: = int 5tantheta*5sectheta*5sec^2thetad theta = 125int tanthetasecthetasec^2thetad theta Then, if we let: u = sectheta du = secthetatanthetad theta …

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 is indicated that y=+sqrt(25-x^2), then the graph is the semi-circle above the x-axis, with dead-end-discontinuations, at (+-5, 0), and there exists this limit problem

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 the same as yours, just factorized differently. $\endgroup$ –