Find the equation of the line that is perpendicular to LaTeX: y=-\frac{1}{5}x+3y = − 1 5 x + 3 and goes through the point (4, -3
1 answer:
Answer:
Step-by-step explanation:
step 1
Find the slope of the line that is perpendicular to the given line
we have
-----> given line
The slope of the given line is
Remember that
If two lines are perpendicular, then their slopes are opposite reciprocal
therefore
The slope of the line that is perpendicular to the given line is
step 2
Find the equation of the line in slope intercept form
we have
substitute
solve for b
therefore
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Applying our power rule gets us our first derivative,
simplifying a little bit,
looking for critical points,
We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.
When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.
Then apply your Zero-Factor Property,
and solve for x in each case to find your critical points.
Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.
Let's come back to this,
and take our second derivative.
Looking for inflection points,
Again, pulling out the smaller power of x, and fractional part,
And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.
Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.
simplifying a little bit,
looking for critical points,
We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.
When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.
Then apply your Zero-Factor Property,
and solve for x in each case to find your critical points.
Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.
Let's come back to this,
and take our second derivative.
Looking for inflection points,
Again, pulling out the smaller power of x, and fractional part,
And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.
Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.
Hi!
y = 16
x = 32
<h3 /><h3>In 30-60-90 triangles, the hypotenuse is double the length of the shortest leg, and the longer leg isWe are given the longest leg. To find the shortest leg, we must divide the longest leg value by
The radicals cancel out and we are left with 16.
The length of y (the shortest leg) is 16.
Now, we also know that the hypotenuse is double the shortest leg. The shortest leg is 16, so if we double that, it's 32.