Write the equation of a line that is perpendicular to the given line and that passes through the given point. 2x + 4y = –6; (2,
1 answer:
First we have to find the slope of the given line. We do that by puttiing the equation into slope-interept form. or solving it for y (same thing). 2x + 4y = -6. We move the 2x over by subtraction to get 4y = -2x - 6. Divide both sides by 4 to get the slope-intercept form we are seeking. . The slope that is perpendicular to that slope of -1/2 is the opposite reciprocal. Therefore, the perpendicular slope is a positive 2/1 or just 2. If this new line goes through the point (2, 5), it goes through the x value of 2 and the y value of 5. We will use those values along with the slope of 2 to solve for b in our slope-intercept equation. That looks like this:
and b = 1. That means our new equation, the one that is perpendicular to the given line, has an equation of y = 2x + 1.
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<h3>Given</h3>
tan(x)²·sin(x) = tan(x)²
<h3>Find</h3>x on the interval [0, 2π)
<h3>Solution</h3>Subtract the right side and factor. Then make use of the zero-product rule.
... tan(x)²·sin(x) -tan(x)² = 0
... tan(x)²·(sin(x) -1) = 0
This is an indeterminate form at x = π/2 and undefined at x = 3π/2. We can resolve the indeterminate form by using an identity for tan(x)²:
... tan(x)² = sin(x)²/cos(x)² = sin(x)²/(1 -sin(x)²)
Then our equation becomes
... sin(x)²·(sin(x) -1)/((1 -sin(x))(1 +sin(x))) = 0
... -sin(x)²/(1 +sin(x)) = 0
Now, we know the only solutions are found where sin(x) = 0, at ...
... x ∈ {0, π}