Question (1):The general formula of the quadratic equation is:
ax² + bx + c = 0
The given equation is:
5x² + 9x = 4
Rearrange the given equation to look the standard one:
5x² + 9x - 4 = 0
Now, compare the coefficients in the given equation with the standard one, you will find that:
a = 5, b = 9 and c = -4
Question (2):The given expression is:
-5 + 2x²<span> = -6x
</span>Rearrange this expression to be in standard form:
2x² + 6x - 5 = 0
This means that:
a = 2
b = 6
c = -5
The roots of the equation can be found using the formula in the attached image.
Substituting in this formula with the given a, b and c, we would find that the correct choice is third one (I have attached the correct choice)
Question (3):Quadratic formula (the one used in the previous question, also shown in attached images) is the best method to get the solution of any quadratic equation. This is because, putting the equation in standard form, we can simply get the values of a, b and c, substitute in the formula and get the precise solutions of the equation.
Hope this helps :)
Step-by-step explanation:
yes its ans is (3,3) fast you have see in x axis then y axis .so according to this we write 3 which lies in x- axis and then another 3 which lies in y-axis.
answer one would be 18 square root of 5
and the second would be 4 square root of 5
Step-by-step explanation:
Answer:
t ∈ {1, 3}
Step-by-step explanation:
You want to find t such that ...
h = 27
27 = -8t^2 +32t +3 . . . . . . substitute the expression for h
24 = -8t^2 +32t . . . . . . . . . subtract 3
-3 = t^2 -4t . . . . . . . . . . . . . divide by -8
1 = t^2 -4t +4 = (t -2)^2 . . . . add 4 to complete the square
±√1 = t -2 . . . . . . . . . . . . . . take the square root
t = 2 ± 1 . . . . . . . . . . . . . . . . add 2
t = 1 or 3
The object is 27 ft off the ground at t = 1 and again at t = 3.
X=1 is the answer to your question