we know the x-intercept of the line is 1, recall that an x-intercept is when the graph intercepts or touches the x-axis, and when that happens, y = 0, so the point is really x = 1, y = 0, namely (1,0). We also know another point on the line, is (-2, 9).

$\bf \stackrel{x-intercept}{(\stackrel{x_1}{1}~,~\stackrel{y_1}{0})}\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{9}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{9-0}{-2-1}\implies \cfrac{9}{-3}\implies -3 \\\\\\ \stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-0=-3(x-1) \\\\\\ ~\hspace{10em}y=-3x+3$

5 0

2 years ago

HELPPPPPPPP!!!!!!!!!!

Novosadov [1.4K]

Step-by-step explanation:

98 = 188t - 16t^2

-16t^2 + 188t = 98

-16t^2 + 188t - 98 = 0

8t^2 - 94t + 49 = 0

Use the quadratic formula with a = 8 , b = -94 , c = 49 to get the solutions

t = [(47 - root 1817) / 8]≈ 0.55

and

t = [(47 + root 1817) / 8] ≈ 11.20

8 0

3 years ago