$\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ h=-16t^2+\stackrel{\stackrel{v_o}{\downarrow }}{65}t$

now, take a look at the picture below, so for 2) and 3) is the vertex of this quadratic equation, 2) is the y-coordinate and 3) the x-coordinate.

$\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+65}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{65}{2(-16)}~~,~~0-\cfrac{65^2}{4(-16)} \right) \implies \left( \cfrac{65}{32}~,~0- \cfrac{4225}{-64}\right)$

$\bf \left( \cfrac{65}{32}~,~0+ \cfrac{4225}{64}\right)\implies \left( \stackrel{seconds}{2\frac{1}{32}}~~,~~ \stackrel{feet~hight}{66\frac{1}{64}}\right)$

6 0

2 years ago

in the xy-plane, the line x + y=k, where k is a constant, is tangent to the graph of y=x^2+3x+1. What is the value of k?

Sholpan [36]

When a line and a curve are tangent to each other, they are joined together by 1 point (x,y). So, you equate both equations either in terms of x or y. For this problem, let's equate in terms of y.

y = k - x
y = x² + 3x + 1

k - x = x² + 3x + 1
k = x² + 3x + x + 1
k = x² + 4x + 1

This would be the value of k. Since we are not given the common point, it can only be expressed in terms of x or y.

3 0

3 years ago

Answer any of these, no work is necessary. <br> Thank you so much!

faust18 [17]

Answer: use math-way Have a nice day

3 0

2 years ago

Prove that<br> cos^4∆ - sin^4∆ = 2cos^2∆ - 1

GarryVolchara [31]

$\cos^{4} x-\sin^{4} x\\\\=(\cos^{2} x+\sin^{2} x)(\cos^{2} x-\sin^{2} x)\\\\=\cos^{2} x-\sin^{2} x\\\\=\cos^{2} x-(1-\cos^{2} x)\\\\=2\cos^{2} x-1$

3 0

1 year ago