Answer:
(b) 21.4
Step-by-step explanation:
There are a couple of interesting relations regarding chords and secants and tangents of a circle. With the right point of view, they can be viewed as variations of the same relation, possibly making them easier to remember.
When chords cross inside a circle (as here), each divides the other into two parts. The product of the lengths of the two parts of one chord is the same as the product of the lengths of the two parts of the other chord.
Here, that means ...
7x = 10·15
x = 150/7 = 21 3/7 ≈ 21.4
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<em>Additional comment</em>
A secant is a line that intersects a circle in two places. (A tangent is a special case of secant where the two points of intersection are the same point.) When two secants meet outside the circle, there is a special relation between the lengths of the various line segments.
Consider the line segment from the point where the secants meet each other to the far intersection point with the circle. The product of that length and the length to the near intersection point with the circle is the same for both secants.
Here's the viewpoint that merges these two relations:
<em>The product of the lengths from the point of intersection of the lines with each other to the two points of intersection with the circle is the same for each line</em>.
(Note that when the "secant" is a tangent, that product is the square of the distance from the tangent point to the point of intersection with the other line--the distance to the circle multiplied by itself.)
3.20 divieded my 2 =1.6 spring 3.20 +1.6 =4.8
The Answer Is 42.............................................
Answer:
A, B, and C
Step-by-step explanation:
a). -6 ÷ 2 x 5 1/3
= -3 x 5 1/3
= -16
b). 24 ÷ (-3) + 1/2
= -8 + 1/2
= -7.5
c). 36 ÷ 4 x (-2)
= 9 x -2
= -18
d). -54 ÷ (-9) x (-3 2/3)
6 x -3 2/3
= -22
RemarkIf you don't start exactly the right way, you can get into all kinds of trouble. This is just one of those cases. I think the best way to start is to divide both terms by x^(1/2)
Step OneDivide both terms in the numerator by x^(1/2)
y= 6x^(1/2) + 3x^(5/2 - 1/2)
y =6x^(1/2) + 3x^(4/2)
y = 6x^(1/2) + 3x^2 Now differentiate that. It should be much easier.
Step TwoDifferentiate the y in the last step.
y' = 6(1/2) x^(- 1/2) + 3*2 x^(2 - 1)
y' = 3x^(-1/2) + 6x I wonder if there's anything else you can do to this. If there is, I don't see it.
I suppose this is possible.
y' = 3/x^(1/2) + 6x
y' =

Frankly I like the first answer better, but you have a choice of both.