![SOLVED: Entered Answer Preview Result (9/[(pi^3)*(n^3)1)*(16/[(pi^3)*(m^3)])* 16 [pitn-sin(pi-n)+2*cos(n-pi)-2]+ Tn sin(Tn) + 2 cos(nr) 2)(rm sin(wm) + 2 cos(wm) 2) incorrect T3n3 r8m? [pitm-sin(pitm)+2*cos(pi*m)-2] The answer above is NOT correct ... SOLVED: Entered Answer Preview Result (9/[(pi^3)*(n^3)1)*(16/[(pi^3)*(m^3)])* 16 [pitn-sin(pi-n)+2*cos(n-pi)-2]+ Tn sin(Tn) + 2 cos(nr) 2)(rm sin(wm) + 2 cos(wm) 2) incorrect T3n3 r8m? [pitm-sin(pitm)+2*cos(pi*m)-2] The answer above is NOT correct ...](https://cdn.numerade.com/ask_images/6d8471db5b624058a255c6631ad86008.jpg)
SOLVED: Entered Answer Preview Result (9/[(pi^3)*(n^3)1)*(16/[(pi^3)*(m^3)])* 16 [pitn-sin(pi-n)+2*cos(n-pi)-2]+ Tn sin(Tn) + 2 cos(nr) 2)(rm sin(wm) + 2 cos(wm) 2) incorrect T3n3 r8m? [pitm-sin(pitm)+2*cos(pi*m)-2] The answer above is NOT correct ...
sin(pi/10) + sin(13pi/10) is equal to (1) 1/2 (2) -1/2 (3) -1/4 (4) 1 - Sarthaks eConnect | Largest Online Education Community
![SOLVED: Entered Answer Preview Result [(pi"[e^ (-pi*s)MV([(s^2)+(pi^2)] [(s^ 2)+9]1]+ [p([(s^2)+(pi^2)]*[(s^2)+911] incorrect + 12 ) (s2 +9) 72 +32) +9) (t-pi) " ([pi(pi^2)]-9)*[(1/3)*sin(3)*(t-pi)- (Ipi) sin(pi)"(t-pi)]+(pi (pi^ 2)-9])*[(1/3)*sin(3t ... SOLVED: Entered Answer Preview Result [(pi"[e^ (-pi*s)MV([(s^2)+(pi^2)] [(s^ 2)+9]1]+ [p([(s^2)+(pi^2)]*[(s^2)+911] incorrect + 12 ) (s2 +9) 72 +32) +9) (t-pi) " ([pi(pi^2)]-9)*[(1/3)*sin(3)*(t-pi)- (Ipi) sin(pi)"(t-pi)]+(pi (pi^ 2)-9])*[(1/3)*sin(3t ...](https://cdn.numerade.com/ask_images/0f6b64aaa71d4953add664ff5382ce76.jpg)
SOLVED: Entered Answer Preview Result [(pi"[e^ (-pi*s)MV([(s^2)+(pi^2)] [(s^ 2)+9]1]+ [p([(s^2)+(pi^2)]*[(s^2)+911] incorrect + 12 ) (s2 +9) 72 +32) +9) (t-pi) " ([pi(pi^2)]-9)*[(1/3)*sin(3)*(t-pi)- (Ipi) sin(pi)"(t-pi)]+(pi (pi^ 2)-9])*[(1/3)*sin(3t ...
![calculus - What is $\lim\limits_{x \to 0} \frac{\sin(\pi \cos^2x)}{x^2}= \ ?$ - Mathematics Stack Exchange calculus - What is $\lim\limits_{x \to 0} \frac{\sin(\pi \cos^2x)}{x^2}= \ ?$ - Mathematics Stack Exchange](https://i.stack.imgur.com/EGjCz.jpg)