Maybe you thought about this:
m^2+6m-77+5=0 => m^2+6m-72=0 => m^2+ 2*3*m+3^2-3^2-72=0
=> m^2+2*3*x+3^2= (m+3)^2 this is the complete square of the binomial, we will still keep it =>
(m+3)^2 -9 -72=0 => (m+3)^2 - 81 = 0 => (m+3)^2 - 9^2 = 0
Now we got the difference squares, it mean a^2 - b^2 = (a-b) (a+b)
In according to this => (m+3-9) (m+3+9) = 0 =>
(m-6) (m+12)=0 => m-6=0 or m+12=0 => m1=6 or m2= -12 !!!!
Are you satisfied with this solution?
Answer:
9+ -8+ -7= -6
-8 + -7= -15
Step-by-step explanation:
As you already written, you have to use Newton's law
where F is the force, m is the mass, and a is the acceleration.
You already know the mass and the force, so you have to solve the equality for the acceleration:
plug the given values:
Ummm I think you should email your teacher because they can always help!!
<span> You have :
----------------
v0 = 35 m/s
B = angle of elevation = 30 deg
a suby = - g
v suby0 = ( v0 ) ( sin B )
v suby0 = ( 35 m/s ) ( sin 30 ) = 17.5 m/s
At maximum height :
------------------------------
v suby = vy* = 0.0
t = t*
y = ymax = y*
Basic kinematics gives :
-------------------------------------
a suby = dv suby / dt = -g ------> dt = dv suby /a suby
v suby = dy/dt ------> dt = dy / vsuby
dt = dv suby / a suby = dy / v suby
( v suby ) ( d v suby) = ( a suby ) dy = ( -g ) dy
Now integrate and get :
-----------------------------------
( v suby* )^2 - ( v suby0)^2 ( 1/2 ) = ( -g ) ( ymax )
ymax = [ ( v suby0 )^2 - ( v suby* )^2 ] / [ ( 2 ) ( g ) ]
ymax = ( 17.5 m/s )^2 - ( 0.0 m/s )^2 ] / [ ( 2 ) ( 9.807 m/s^2 ) ]
ymax = 15.6 m <------------------
Check the result:
--------------------------
t* = v suby0 / g = ( 17.5 m/s ) / ( 9.807 m/s^2 ) = 1.784 s
y* - y0 = ( v suby0 ) ( t* ) - ( g/2 ) ( t* )^2
y* = ( 17.5 m/s ) ( 1.784 s ) - ( 9.807 m/s^2 / 2 ) ( 1.784 s )^2
y* = ymax = 31.23 m - 15.61 m = 15.6 m
The ymax values agree.</span>
