Answer:
64
Step-by-step explanation:
<em>[using calculus] </em>When the function h(t) reaches its maximum value, its first derivative will be equal to zero (the first derivative represents velocity of the ball, which is instantaneously zero). We have 
, which equals zero when 
. The ball therefore reaches its maximum height when t = 1.5. To find the maximum height, we need to find h(1.5), which is 64 feet.
<em>[without calculus] </em>This is a quadratic function, so its maximum value will occur at its vertex. The formula for the x-coordinate of the vertex is -b/2a, so the maximum value occurs when t = -48/(2*16), which is 1.5. The maximum height is h(1.5), which is 64 feet.
Consider a homogeneous machine of four linear equations in five unknowns are all multiples of 1 non-0 solution. Objective is to give an explanation for the gadget have an answer for each viable preference of constants on the proper facets of the equations.
Yes, it's miles true.
Consider the machine as Ax = 0. in which A is 4x5 matrix.
From given dim Nul A=1. Since, the rank theorem states that
The dimensions of the column space and the row space of a mxn matrix A are equal. This not unusual size, the rank of matrix A, additionally equals the number of pivot positions in A and satisfies the equation
rank A+ dim NulA = n
dim NulA =n- rank A
Rank A = 5 - dim Nul A
Rank A = 4
Thus, the measurement of dim Col A = rank A = five
And since Col A is a subspace of R^4, Col A = R^4.
So, every vector b in R^4 also in Col A, and Ax = b, has an answer for all b. Hence, the structures have an answer for every viable preference of constants on the right aspects of the equations.
I'll show a step-by-step. I'm not in the mood to explain right now lol.
2(x-2.6)+2.91=7.71
2x-5.2+2.91=7.71
+5.20 +5.20
2x+2.91=12.91
-2.91 -2.91
2x=10
/2 /2
x=5
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hope it helps
sorry if it doesn't
It is in simple form. There is absolutely nothing more that can be done, unless it is a joke. And math jokes are very hard to come by.
I don't think that y(y - 1)/(a + 2b) is any simpler.
Come to think of it, is that x^2 or y^2 on the top left.
I think my first answer is the best one. Nothing can be done.
