- #1
Zashmar
- 48
- 0
15=35 sin Theta*t -4.9t^2
I have tried solving for t, I do not know how to...
please help
thank you
I have tried solving for t, I do not know how to...
please help
thank you
It doesn't look like you applied the formula correctly. What did you use for a, b, and c?Zashmar said:So...
x= [itex](4.9 \pm \sqrt{24.01+2100 sin\vartheta}/(70 sin \vartheta)[/itex]
You need to get the algebra right before we can help you simplify anything.Zashmar said:So I have it now, how would i simplify the following (I have subbed it into x=35cos[itex]\vartheta*t[/itex])?
x=35cos[itex]\vartheta[/itex]([itex]-35 sin\vartheta \pm \sqrt{(35sin \vartheta)^2 -294}/-9.8[/itex]
I tried to put in the parentheses but it would not let me
Chestermiller said:You need to get the algebra right before we can help you simplify anything.
A quadratic equation is an equation of the form ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable. It is a type of polynomial equation and the highest power of the variable is 2.
To solve a quadratic equation for T, you can use the quadratic formula, which is T = (-b±√(b^2-4ac))/2a. First, determine the values of a, b, and c from the given equation and then substitute them into the formula to find the two values of T.
No, a quadratic equation can have at most two solutions for T. This is because a quadratic equation can have at most two real roots, and the solutions for T are the values of x that make the equation equal to 0.
A quadratic equation has real solutions if the discriminant (b^2-4ac) is greater than or equal to 0. If the discriminant is negative, the equation does not have any real solutions.
Yes, there are other methods for solving quadratic equations such as factoring, completing the square, and graphing. However, the quadratic formula is the most reliable and efficient method for finding the solutions to a quadratic equation.