If the 2 lines are parallel, their slopes are the same.
The slope of both is
3+3
m = ---------- = 3
2 - 0
If the new line <span>passes through the point (−1, −1), then the equation in point slope form is:
y+1 = 3(x+1).</span>
Answer:
A exterior angle of a polygon is an angle outside the polygon formed by one of its sides and the extension of an adjacent side.
Let
be the length of the rectangle and
be the width. In the problem it is given that
. It is also given that the area
. Substituting in the length in terms of width, we have
. Using the zero product property,
. Solving these we get the width
. However, it doesn't make sense for the width to be negative, so the width must be
. From that we can tell the length
.
So it tells us that g(3) = -5 and g'(x) = x^2 + 7.
So g(3) = -5 is the point (3, -5)
Using linear approximation
g(2.99) is the point (2.99, g(3) + g'(3)*(2.99-3))
now we just need to simplify that
(2.99, -5 + (16)*(-.01)) which is (2.99, -5 + -.16) which is (2.99, -5.16)
So g(2.99) = -5.16
Doing the same thing for the other g(3.01)
(3.01, g(3) + g'(3)*(3.01-3))
(3.01, -5 + 16*.01) which is (3.01, -4.84)
So g(3.01) = -4.84
So we have our linear approximation for the two.
If you wanted to, you could check your answer by finding g(x). Since you know g'(x), take the antiderivative and we will get
g(x) = 1/3x^3 + 7x + C
Since we know g(3) = -5, we can use that to solve for C
1/3(3)^3 + 7(3) + C = -5 and we find that C = -35
so that means g(x) = (x^3)/3 + 7x - 35
So just to check our linear approximations use that to find g(2.99) and g(3.01)
g(2.99) = -5.1597
g(3.01) = -4.8397
So as you can see, using the linear approximation we got our answers as
g(2.99) = -5.16
g(3.01) = -4.84
which are both really close to the actual answer. Not a bad method if you ever need to use it.