Find slope of line A:
Move into slope-intercept form y = mx+b
<span>5x + 8y = -9
8y = -5x - 9
y = (-5/8)x - 9/8
The slope of line A is -5/8.
If </span><span>Line B is perpendicular to line A, then
slope Line B = negative reciprocal of slope Line A</span>
<span>slope Line B = 8/5
So like B has the equation
y = (8/5)x + b
If it passes through (10,10), we know that when x = 10, y = 10. Use those values to solve for b:
</span>
<span>y = (8/5)x + b
10 = (8/5)·10 + b</span>
<span>10 = (8)·2 + b
10 = 16 + b
b = -6
So line B has equation </span>
<span>y = (8/5)x - 6
m = 8/5 and b = -6
so
m + b = 8/5 - 6 = 8/5 - 30/5 = -22/5
So m+b = -22/5 or -4.4 in decimal form
</span>
Answer:
Infinity
Step-by-step explanation:
It is possible for the coin to endlessly fall on tails, so there are infinity different outcomes.
Tell me if I'm wrong :)
The <span>isosceles triangle has two congruent sides
Their lengths are (</span> x + 3.8 ) and <span>16
Equate them :
x+3.8=16
Solve for x:
3=16-3.8=12.2
</span>
bearing in mind that standard form for a linear equation means
• all coefficients must be integers, no fractions
• only the constant on the right-hand-side
• all variables on the left-hand-side, sorted
• "x" must not have a negative coefficient
![\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-7=1[x-(-1)]\implies y-7=x+1 \\\\\\ y=x+8\implies \boxed{-x+y=8}\implies \stackrel{\textit{standard form}}{x-y=-8}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-7%3D1%5Bx-%28-1%29%5D%5Cimplies%20y-7%3Dx%2B1%20%5C%5C%5C%5C%5C%5C%20y%3Dx%2B8%5Cimplies%20%5Cboxed%7B-x%2By%3D8%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bstandard%20form%7D%7D%7Bx-y%3D-8%7D)
just to point something out, is none of the options, however -x + y = 8, is one, though improper.
