Two circles<span> of </span>radius<span> 4 are </span>tangent<span> to the </span>graph<span> of y^</span>2<span> = </span>4x<span> at the </span>point<span> (</span>1<span>, </span>2<span>). ... I know how to </span>find<span> the </span>tangent<span> line from a circle and a given </span>point<span>, but ... </span>2a2=42. a2=8. a=±2√2. Then1−xc=±2√2<span> and </span>2−yc=±2√2. ... 4 from (1,2<span>), so you could </span>find these<span> centers, and from there the</span>equations<span> of the circle
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Answer:
by doing it the same way you did with the first one.
Step-by-step explanation:
Answer: c
Step-by-step explanation:
bc
To solve this question, you would need to find the common denominator among the 2 fractions. In this case it would be
72.
Then multiply numbers to both of the fractions to get to the bottom number in both fractions equaling 72.
40/72 + 27/72 =67/72.
Try to reduce to lowest terms by dividing a common number from both of them if possible.
When the direction is reversed velocity = 0
v = ds/dt = 6t^2 - 42t + 60 = 0
t^2 - 7t + 10 = 0
(t - 5)(t - 2) = 0
t = 2 s and 5 s which are the times when the direction is reversed
Position at t = 2 = 2(2)^3 - 21(2)^2 +60(2) + 3 = 55 feet
at t = 5 position = 28 feet
Acceleration = dv/dt = 12t - 42
At t = 2 acceleration = -18 ft s-2
At t = 5 ............... = 18 ft s-2
