49t^2 - 16 = 0
show work
2006-09-09 12:19:55 · 8 answers · asked by Kevin 5 in Science & Mathematics ➔ Mathematics
I would show you easy way
just do what i do;
49t^2 - 16 = 0
49t^2 = 16
√(49t^2) = √16
7t = +4 ; t = + 7/4
7t = - 4 ; t = - 7/4
you see . nice and easy.
Good Luck & Good Question.
2006-09-09 12:54:19 · answer #1 · answered by sweetie 5 · 0⤊ 0⤋
Use the quadratic formula, which is: -b(+ and -)*sqroot(b^2-4ac) all divided by 2a.
Where 49t^2 is "a", there is no "b" and "c" is 16.
Another way is to expand the equation.
(7t-4)(7t + 4)=0. Therefore t=4/7 or t= -4/7
2006-09-09 19:44:32 · answer #2 · answered by Axel 2 · 0⤊ 1⤋
Pascal is correct
49t^2 = 16
t^2 = 16/49
t = (16/49)^0.5
Since the rule of multiplication is + x + and - x - are both positive
t = + or - (4/7)
Can check by substitution
49 (4/7)^2 =16
49*16/49 = 16
49's cancel (since 49/49 =1) so 16 = 16
2006-09-09 19:33:34 · answer #3 · answered by happyboy 2 · 0⤊ 1⤋
49t²-16=0
49t²=16
t²=16/49
t=屉(16/49)
t=±4/7
2006-09-09 19:23:21 · answer #4 · answered by Pascal 7 · 2⤊ 0⤋
6
2006-09-09 19:20:45 · answer #5 · answered by mae mae 3 · 0⤊ 3⤋
49(t)^2-16=0
49(t)^2=16
t^2=16/49
t=(16/49)^ -2
t=[(4*4)/(7*7)]* -2
t=4/7
2006-09-09 19:24:14 · answer #6 · answered by Toshika 2 · 1⤊ 1⤋
this sure looks like a difference of squares problem
difference of squares is one of the little algebra signals a high school math student should always be on the look out for
both terms are obviously squares and they are subtracted
factor like difference of squares
(7t-4)(7t+4)=0
solve each root
7t-4=0
7t=4
t=4/7
and
t=-4/7
woohooo
2006-09-09 19:23:12 · answer #7 · answered by enginerd 6 · 1⤊ 1⤋
t=0 because anything times 0 is 0!
2006-09-09 19:21:28 · answer #8 · answered by Deauxe 3 · 0⤊ 3⤋