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This is in the fourth quadrant where the cosine is positive
By pythagoras the hypotenuse is 5
so the cosine is 3/5 B
Answer:
.
Step-by-step explanation:
The equation of a circle of radius
centered at
is:
.
.
Differentiate implicitly with respect to
to find the slope of tangents to this circle.
![\displaystyle \frac{d}{dx}[x^{2} + y^{2}] = \frac{d}{dx}[25]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bx%5E%7B2%7D%20%2B%20y%5E%7B2%7D%5D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B25%5D)
.
Apply the power rule and the chain rule. Treat
as a function of
,
.
.
.
That is:
.
Solve this equation for
:
.
The slope of the tangent to this circle at point
will thus equal
.
Apply the slope-point of a line in a cartesian plane:
, where
is the gradient of this line, and
are the coordinates of a point on that line.
For the tangent line in this question:
,
.
The equation of this tangent line will thus be:
.
That simplifies to
.
If the value of a is negative, then the range will be (-∞, k) and if the value of the a is positive then the range will be (k, ∞).
<h3>What is a quadratic equation?</h3>
It's a polynomial with a worth of nothing.
There exist polynomials of variable power 2, 1, and 0 terms.
A quadratic condition is a condition with one explanation where the degree of the equation is 2.
Domain and range of linear and quadratic functions
Let the linear equation be y = mx + c.
Then the domain and the range of the linear function are always real.
Let the quadratic equation will be in vertex form.
y = a(x - h)² + k
Then the domain of the quadratic function will be real.
If the value of a is negative, then the range will be (-∞, k) and if the value of the a is positive then the range will be (k, ∞).
More about the quadratic equation link is given below.
brainly.com/question/2263981
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