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

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 …

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

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5月2日 · The answer is (x+5)(x-5) . (x^2-25) fits the pattern of the difference of squares in which (a^2-b^2)=(a+b)(a-b).

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 …

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. $\endgroup$ –

$\begingroup$ Oh thank you, should I write the integral from -4 to 4 of $25-x^2 -3^2$ $\endgroup ...

2018年7月29日 · I recognize the two difference of squares: $49-x^2$ and $25-x^2$. I squared the equation to get: ${49-x^2}-2(\\sqrt{(49-x^2)(25-x^2)})+{25-x^2}=9$ However, I can't quite figure out how to remove the

2021年4月5日 · For a circle with radius of $5$, we know it is, $$x^2 + y^2 = 25.$$ I can interpret it as $f(x,y) = x^2+y^2$ and $g(z)=25$, the final circle is when the two graph ...