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2 answers:
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Similar polygons only differ by a scaling factor. In other words, two polygons are similar if one is the scaled version of the other.
In particular, this implies that the angles are preserved, and the correspondent sides are in proportion.
These two polygons are both rectangles, so the angles are preserved. We must check the sides, and we have to check if the smaller sides are in the same proportion as the bigger sides.
So, the two rectangles are similar if the following is true.
In any proportion, the product of the inner terms must be the same as the product of the outer terms:
This is clearly false, and thus the two rectangles are not similar.
Answer:
Match each of the trigonometric expressions below with theequivalent non-trigonometric function from the following list.Enter the appropiate letter(A,B, C, D or E)in each blank
A . tan(arcsin(x/8))
B . cos (arsin (x/8))
C. (1/2)sin (2arcsin (x/8))
D . sin ( arctan (x/8))
E. cos (arctan (x/8))
These are the spaces to fill out :
.. ..........x/64 (sqrt(64-x^2))
.............x/sqrt(64+x^2)
.............sqrt(64-x^2)/8
..............x/sqrt(64-x^2)
..............8/sqrt(64+x^2)
A. ........tan(arcsin(x/8)) =......x/sqrt(64-x^2)
B . cos (arsin (x/8)) ....sqrt(64-x^2)/8
Step-by-step explanation:
To solve this we have to find the missing sides to each of the triange discribed in prenthesis thus
A we have the sides of the triangle given by x, 8 and or
thus tan(arcsin(x/8)) = =
Therefore ........tan(arcsin(x/8)) =......x/sqrt(64-x^2)
B
Here we have cos = adjacent/hypotenuse where adjacent side is and hypothenuse = 8 we have
/8
B . cos (arsin (x/8)) ....sqrt(64-x^2)/8
