Step-by-step explanation:
after I answered the other one(s), you should be really able to do this yourself.
it is precisely the same method and formula just with different numbers.
what's the problem now ?
y = -16x² + 247x + 141
we assume, the ground is at 0 ft.
so, we need to solve
0 = -16x² + 247 x + 141
the general solution for such quadratic equations is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = -16
b = 247
c = 141
x = (-247 ± sqrt(247² - 4×-16×141))/(2×-16) =
= (-247 ± sqrt(61009 + 9024))/-32 =
= (-247 ± sqrt(70033))/-32
x1 = (-247 + 264.6374879...)/-32 = -0.551171497... s
x2 = (-247 - 264.6374879...)/-32 = 15.9886715... s
again, the negative solution for time did not make any sense, so, x2 is our solution.
the rocket will hit the ground after about 15.99 seconds.
Answer:
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Step-by-step explanation:
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Answer:
Step-by-step explanation:
An eigenvalue of n × n is a function of a scalar 
considering that there is a solution (i.e. nontrivial) to an eigenvector x of Ax =
Suppose the matrix ![A = \left[\begin{array}{cc}-1&-1\\2&1\\ \end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-1%26-1%5C%5C2%261%5C%5C%20%5Cend%7Barray%7D%5Cright%5D)
Thus, the equation of the determinant (A - 1) = 0
This implies that:
![\left[\begin{array}{cc}-1-\lambda &-1\\2&1- \lambda\\ \end{array}\right] =0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-1-%5Clambda%20%26-1%5C%5C2%261-%20%5Clambda%5C%5C%20%5Cend%7Barray%7D%5Cright%5D%20%3D0)
Hence, the eigenvalues of the equation are 
Also, the eigenvalues can be said to be complex numbers.
(x - a)^2 + (y - b)^2 = r^2 is a circle with centre at (a, b) and radius of r
The correct answer is (x - 2)^2 + (y + 10)^2 = 9 ie the first one
