Answer:
the equation of the line that is perpendicular to y = 1/2x + 3 and passes through the point (10, -5)
= -5 = -2x + 15
Step-by-step explanation:
Write an equation of the line that is perpendicular to y = 1/2x + 3 and passes through the point (10, -5).
Using the slope intercept equation,
y = mx +c
m = slope = 1/2
For two lines to be perpendicular, the product of their slopes is -1
Let the slope of the other line be m2
m1×m2 =-1
1/2×m2 = -1
m2 = -1/(1/2) = -2
Slope of line = -2
For points (10, -5), x = 10, y =-5
-5 = -2× 10 +c
-5 = -20+ c
c = -5+20= 15
the equation of the line that is perpendicular to y = 1/2x + 3 and passes through the point (10, -5)
-5 = -2x + 15
you have a quadratic equation that can be factored, like x2+5x+6=0.This can be factored into(x+2)(x+3)=0.
So the solutions are x=-2 and x=-3.
2.
<span><span>1. Try first to solve the equation by factoring. Be sure that your equation is in standard form (ax2+bx+c=0) before you start your factoring attempt. Don't waste a lot of time trying to factor your equation; if you can't get it factored in less than 60 seconds, move on to another method.
</span><span>2. Next, look at the side of the equation containing the variable. Is that side a perfect square? If it is, then you can solve the equation by taking the square root of both sides of the equation. Don't forget to include a ± sign in your equation once you have taken the square root.
3.</span>Next, if the coefficient of the squared term is 1 and the coefficient of the linear (middle) term is even, completing the square is a good method to use.
4.<span>Finally, the quadratic formula will work on any quadratic equation. However, if using the formula results in awkwardly large numbers under the radical sign, another method of solving may be a better choice.</span></span>
Answer:
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Step-by-step explanation:
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