First, find the midpoint of the diameter. the midpoint is the center of the circle.
[(-10+(-6)]/2=-8, [(-8+(-2)]/2=-5
so the center is at (-8, -5)
next, find the distance between the two points:
d=√[-6-(-10)]²+[-2-(-8)]²=√4²+6²=√52=2√13
half of the diameter is radius, r=half of 2√13=√13
standard equation for the circle is (x-h)²+(y-k)²=r², (h,k) being the center, r the radius
(x+8)²+(y+5)²=13 is the answer.
X^2=16
x= 4 OR x=-4
The picture attached below is your answer because it’s between 4 and -4.
Answer:
y = 5x
Step-by-step explanation:
The equation for direct variation is y = kx, where k is the factor of variation. So we know that y and x must fit into this formula.
They've given us a y and x value to solve for k.
y = kx
15 = 3k
15/3 = k
<u>k = 5</u>
Plug k back into our equation.
y = 5x
Answer:
- g(2.95) ≈ -1.8; g(3.05) ≈ -0.2
- A) tangents are increasing in slope, so the tangent is below the curve, and estimates are too small.
Step-by-step explanation:
(a) The linear approximation of g(x) at x=b will be ...
g(x) ≈ g'(b)(x -b) +g(b)
Using the given relations, this is ...
g'(3) = 3² +7 = 16
g(x) ≈ 16(x -3) -1
Then the points of interest are ...
g(2.95) ≈ 16(2.95 -3) -1 = -1.8
g(3.05) ≈ 16(3.05 -3) -1 = -0.2
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(b) At x=3, the slope of the curve is increasing, so the tangent lies below the curve. The estimates are too small. (Matches description A.)