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The equation of a line passing through point (4, -1) and perpendicular to the line whose equation is 2x - y - 7 = 0 is y = -1/2x + 1
<h3>Equation of a line</h3>
A line is the shortest distance between two points. The equation of a line in point-slope form and perpendicular to a line is given as;
y - y1 = -1/m(x-x1)
where
m is the slope
(x1, y1) is the intercept
Given the following
Point = (4, -1)
Line: 2x-y - 7 = 0
Determine the slope
-y = -2x + 7
y= 2x - 7
Slope = 2
Substitute
y+1 = -1/2(x -4)
Write in slope-intercept form
2(y + 1) = -(x - 4)
2y+2 = -x + 4
2y = -x + 2
y = -1/2 + 1
Hence the equation of a line passing through point (4, -1) and perpendicular to the line whose equation is 2x - y - 7 = 0 is y = -1/2x + 1
Learn more on equation of a line here: brainly.com/question/13763238
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Domain means the values of independent variable(input) which will give defined output to the function.
Given:
The height h of a projectile is a function of the time t it is in the air. The height in feet for t seconds is given by the function
Solution:
To get defined output, the height h(t) need to be greater than or equal to zero. We need to set up an inequality and solve it to find the domain values.
![To \; find \; domain:\\\\h(t) \geq0\\\\-16t^2+96t \geq 0\\Factoring \; -16t \; in \; the \; left \; side \; of \; the \; inequality\\\\-16t(t-6) \geq 0\\Step \; 1: Find \; Boundary \; Points \; by \; setting \; up \; above \; inequality \; to \; zero.\\\\t(t-6)=0\\Use \; zero \; factor \; property \; to \; solve\\\\t=0 \; (or) \; t = 6\\\\Step \; 2: \; List \; the \; possible \; solution \; interval \; using \; boundary \; points\\(- \infty,0], \; [0, 6], \& [6, \infty)](https://tex.z-dn.net/?f=%20To%20%5C%3B%20find%20%5C%3B%20domain%3A%5C%5C%5C%5Ch%28t%29%20%5Cgeq0%5C%5C%5C%5C-16t%5E2%2B96t%20%5Cgeq%20%200%5C%5CFactoring%20%5C%3B%20-16t%20%5C%3B%20in%20%5C%3B%20the%20%5C%3B%20left%20%5C%3B%20side%20%5C%3B%20of%20%5C%3B%20the%20%5C%3B%20inequality%5C%5C%5C%5C-16t%28t-6%29%20%5Cgeq%20%200%5C%5CStep%20%5C%3B%201%3A%20Find%20%5C%3B%20Boundary%20%5C%3B%20Points%20%5C%3B%20by%20%5C%3B%20setting%20%5C%3B%20up%20%5C%3B%20above%20%5C%3B%20inequality%20%5C%3B%20to%20%5C%3B%20zero.%5C%5C%5C%5Ct%28t-6%29%3D0%5C%5CUse%20%5C%3B%20zero%20%5C%3B%20factor%20%5C%3B%20property%20%5C%3B%20to%20%5C%3B%20solve%5C%5C%5C%5Ct%3D0%20%5C%3B%20%28or%29%20%5C%3B%20t%20%3D%206%5C%5C%5C%5CStep%20%5C%3B%202%3A%20%5C%3B%20List%20%5C%3B%20the%20%5C%3B%20possible%20%20%5C%3B%20solution%20%5C%3B%20interval%20%5C%3B%20using%20%5C%3B%20boundary%20%5C%3B%20points%5C%5C%28-%20%5Cinfty%2C0%5D%2C%20%5C%3B%20%5B0%2C%206%5D%2C%20%5C%26%20%5B6%2C%20%5Cinfty%29%20)
![Step \; 3:Pick \; test \; point \; from \; each \; interval \; to \; check \; whether \\\; makes \; the \; inequality \; TRUE \; or \; FALSE\\\\When \; t = -1\\-16(-1)(-1-6) \geq 0\\-112 \geq 0 \; FALSE\\(-\infty, 0] \; is \; not \; solution\\Also \; Logically \; time \; t \; cannot \; be \; negative\\\\When \; t = 1\\-16(1)(1-6) \geq 0\\80 \geq 0 \; TRUE\\ \; [0, 6] \; is \; a \; solution\\\\When \; t = 7\\-16(7)(7-6) \geq 0\\-112 \geq 0 \; FALSE\\ \; [6, -\infty) \; is \; not \; solution](https://tex.z-dn.net/?f=%20Step%20%5C%3B%203%3APick%20%5C%3B%20test%20%5C%3B%20point%20%5C%3B%20from%20%5C%3B%20each%20%5C%3B%20interval%20%5C%3B%20to%20%5C%3B%20check%20%5C%3B%20whether%20%5C%5C%5C%3B%20makes%20%5C%3B%20the%20%5C%3B%20inequality%20%5C%3B%20TRUE%20%5C%3B%20or%20%5C%3B%20FALSE%5C%5C%5C%5CWhen%20%5C%3B%20t%20%3D%20-1%5C%5C-16%28-1%29%28-1-6%29%20%5Cgeq%20%200%5C%5C-112%20%5Cgeq%20%200%20%5C%3B%20FALSE%5C%5C%28-%5Cinfty%2C%200%5D%20%5C%3B%20is%20%5C%3B%20not%20%5C%3B%20solution%5C%5CAlso%20%5C%3B%20Logically%20%5C%3B%20time%20%5C%3B%20t%20%5C%3B%20cannot%20%5C%3B%20be%20%5C%3B%20negative%5C%5C%5C%5CWhen%20%5C%3B%20t%20%3D%201%5C%5C-16%281%29%281-6%29%20%5Cgeq%20%200%5C%5C80%20%5Cgeq%20%200%20%5C%3B%20TRUE%5C%5C%20%5C%3B%20%5B0%2C%206%5D%20%5C%3B%20is%20%5C%3B%20a%20%5C%3B%20solution%5C%5C%5C%5CWhen%20%5C%3B%20t%20%3D%207%5C%5C-16%287%29%287-6%29%20%5Cgeq%20%200%5C%5C-112%20%5Cgeq%20%200%20%5C%3B%20FALSE%5C%5C%20%5C%3B%20%5B6%2C%20-%5Cinfty%29%20%5C%3B%20is%20%5C%3B%20not%20%5C%3B%20solution%20)
Conclusion:
The domain of the function is the time in between 0 to 6 seconds
The height will be positive in the above interval.
The answer is no solution
Easiest way is to find which one is smaller
-2 is smaller than 6
8 is smaler than 10
so (-2,8) is smaller
we find the distance betewen them
distance between -2 and 6 is 8
distance beteeen 8 and 10=2
so
we add 1/2 of each to the smaller one
8/2=4
-2+4=2
2/2=1
8+1=9
the midpoint is (1,9)
