3 years ago

H(t)=- 16t^2 +16t+12 zeros of the parabola

Mathematics

1 answer:

igomit [66]3 years ago

8 0

Step-by-step explanation:

h(t) = -16t² + 16t + 12

h(t) = -4(4t² - 4t - 3)

h(t) = -4(2t - 3)(2t + 1)

When h(t) = 0, we have t = 1.5 or t = -0.5.

Hence the zeroes are 1.5 and -0.5.

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Read 2 more answers

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zloy xaker [14]

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$x=-\dfrac{44}{7}$

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Step-by-step explanation:

Arrange the equations so that all the variables are in columns (x as the first column, y as the second, equals a number as the third):

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Define and edit Matrix A. We need a 3 x 2 matrix. Input the coefficients and constant:

$\left[\begin{array}{ccc}-11&-4&36\\5&5&-10\end{array}\right]$

Find rref function, enter matrix A: rref([A]) (press enter)

This gives you:

$\left[\begin{array}{ccc}1&0&-\frac{44}{7}\\0&1&\frac{58}{7}\end{array}\right]$

which means:

$x=-\dfrac{44}{7}$

$y=\dfrac{58}{7}$

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