Answer this in 30 seconds for brainiests
2 answers:
You might be interested in
Answer:
A
Step-by-step explanation:
The Tangent-Secant Exterior Angle Measure Theorem states that if a tangent and a secant or two tangents/secants intersect outside of a circle, then the measure of the angle formed by them is half of the difference of the measures of its intercepted arcs. Basically, what that means here is that equals half of the difference of
and the measure of the unlabeled arc.
First, we need to find the measure of the unlabeled arc, since we can't find without it. We know that the measure of the full arc formed by the circle is
, so the measure of the unlabeled arc must be
by the Arc Addition Postulate.
Now, we can find . Using all of the information known, we can solve for
like this:
Hope this helps!
The given equations are
(1)
(2)
When t=0, obtain
Obtain derivatives of (1) and find x'(0).
x' (t+1) + x - 4√x - 4t*[(1/2)*1/√x = 0
x' (t+1) + x - 4√x -27/√x = 0
When t=0, obtain
x'(0) + x(0) - 4√x(0) = 0
x'(0) + 9 - 4*3 = 0
x'(0) = 3
Here, x' means
.
Obtain the derivative of (2) and find y'(0).
2y' + 4*(3/2)*(√y)*(y') = 3t² + 1
When t=0, obtain
2y'(0) +6√y(0) * y'(0) = 1
2y'(0) = 1
y'(0) = 1/2.
Here, y' means
.
Because
, obtain
Answer:
The slope of the curve at t=0 is 1/6.
When t=0, obtain
Obtain derivatives of (1) and find x'(0).
x' (t+1) + x - 4√x - 4t*[(1/2)*1/√x = 0
x' (t+1) + x - 4√x -27/√x = 0
When t=0, obtain
x'(0) + x(0) - 4√x(0) = 0
x'(0) + 9 - 4*3 = 0
x'(0) = 3
Here, x' means
Obtain the derivative of (2) and find y'(0).
2y' + 4*(3/2)*(√y)*(y') = 3t² + 1
When t=0, obtain
2y'(0) +6√y(0) * y'(0) = 1
2y'(0) = 1
y'(0) = 1/2.
Here, y' means
Because
Answer:
The slope of the curve at t=0 is 1/6.