2t+3 = -11 Subtract 3 from both sides: 2t+3-3 = -11-3 2t = -14 Divide both sides by 2 in order to find the value of t: t = -7
5t + 2 = 3t - 2t + 5 Collect like terms 5t - 3t + 2t = 5 - 2 4t = 3 t = ¾ Check: 5 x 3/4 = 3¾ + 2 = 5¾; 3x¾- 2x¾ + 5 = 5¾ QED
3tan (2t) = sqrt(3) tan (2t) = sqrt(3)/3 = 0.577 from tangent tables, 2t = 30 degrees t = 15 degrees
T+(3-2t)=4t+1. Assuming the two T's , T and t are the same variable, lets proceed. t+(3-2t)=4t+1 3-t=4t+1 2-t=4t 2=5t last step is yours
6t - 5t - 6 = (3t + 2)(2t - 3)
2t+3 = -11 Subtract 3 from both sides: 2t+3-3 = -11-3 2t = -14 Divide both sides by 2 in order to find the value of t: t = -7
2t-3-3 = 7 2t = 7+6 2t = 13 t = 6.5
Subtract 3 from each side: 2t = -11 - 3 Divide each side by 2: t = -14/2 = -7
2t^2+5t-3=0 (2t-1)(t+3)=0 2t-1=0 and t+3=0 t=.5 and t=-3
2t + 3 = -9Subtract 3 from each side:2t = -12Divide each side by 2:t = -6======================-3t - 7 = (-3)(-6) - 7 = +18 - 7= 11.
Below ^ denotes power. 8q^6r^3=(2q^2r)^3 denote as a^3 and find a=2q^2r 27s^6t^3=(3s^2t)^3 denote as b^3 and find b=3s^2t Now 8q^6r^3+27s^6t^3=a^3+b^3 = (a+b)(a^2-ab+b^2) | substitute back =(2q^2r+3s^2t)(4q^4r^2-6q^2rs^2t+9s^4t^2). Notice (2q^2r)^2=4q^4r^2 (3s^2t)^2=9s^4t^2. Hence 8q^6r^3+27s^6t^3=(2rq^2+3ts^2)(4r^2q^4-6rts^2t^2+9t^2s^4), a=2q^2r and b=3s^2t.
5t + 2 = 3t - 2t + 5 Collect like terms 5t - 3t + 2t = 5 - 2 4t = 3 t = ¾ Check: 5 x 3/4 = 3¾ + 2 = 5¾; 3x¾- 2x¾ + 5 = 5¾ QED
3tan (2t) = sqrt(3) tan (2t) = sqrt(3)/3 = 0.577 from tangent tables, 2t = 30 degrees t = 15 degrees
T+(3-2t)=4t+1. Assuming the two T's , T and t are the same variable, lets proceed. t+(3-2t)=4t+1 3-t=4t+1 2-t=4t 2=5t last step is yours
t=3/2 or t=2 1/2
(t + 1)(2t - 3)
12(t+6)+8=4(2t-1)+2t 12t+72+8=8t-4+2t 12t-2t-8t=-72-8-4 2t=-84 2t/2=84/2 t=42