If you take a moment and solve each equation (find the value
of the letter in each one), it'll jump out at you.
Here, let me do it for you. Why should you tire yourself.
5x = 20 . . . . . x = 4
4b = 7 . . . . . . b = 7/4
8w = 32 . . . . . w = 4
12y = 48 . . . . . y = 4
The second equation is not like the others.
It's the only one where the letter is not equal to 4 .
Answer:
t = 2
Step-by-step explanation:
Notice that this expression for the projectile's path is that of a quadratic function with negative leading term. The graph of it therefore consists of a parabola with the branches pointing down (due to he negative leading coefficient). Therefore, the maximum of such parabola will reside at its vertex.
Recall that the formula for the position of the vertex in a general parabolic function of the form:
, is given by the expression: 
In our case, the variable "x" is in fact "t", the leading coefficient (
) is -5, and the coefficient for the linear term (
) is 20.
Therefore, the maximum of the path will be when 
The solution for that proplem is X=4,1
0.023
if u divide 23 by one thousand you get 23 thousandths as a decimal