2y^3 – 2y – 10y + 10 + y^2 – 1 < 0 [the terms are simply reorganized again]
factor 2y from the first two terms, -10 from the second two terms
2y (y^2 -1) – 10 (y-1) + y^2 – 1 < 0
2y (y+1)(y–1) – 10 (y-1) + (y+1)(y–1) < 0 [ because y^2 – 1 = (y+1)(y–1) ]
factor out (y-1) from all the terms
(y-1) [2y(y+1)-10+ y+1] < 0
(y-1) [(y+1) (2y+1) - 10] < 0
Let us simplify (y+1) (2y+1) - 10 < 0 now
(y-1) (2y^2+y+2y+1-10) < 0
(y-1) (2y^2 +3y -9 < 0
(y-1) (2y^2 +6y -3y - 9) < 0 [ because 3y = 6y -3y] j
Answer:
8.3
Step-by-step explanation:
<em><u>Given</u></em><em><u>, </u></em>
Circumference of the circle = 52 cm
<em><u>Therefore</u></em><em><u> </u></em><em><u>,</u></em>
<em>By</em><em> </em><em>the</em><em> </em><em>problem</em><em>, </em>



=> r = 8.2802547771
<em>Rounded</em><em> </em><em>to</em><em> </em><em>nearest</em><em> </em><em>tenth</em><em>,</em>
=> <em><u>r = 8.3 (Ans)</u></em>
Answer:
A) Yes, because the sum of any two side lengths is greater than the third side length.
Step-by-step explanation:
All triangles have a property that the sum of two side lengths is greater than the remaining side length.
4 + 8 = 12 > 10, making it a triangle

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Answer:

Step-by-step explanation:
