The answer is 0.6 the 6's go on forever
Answer:
Alternate exterior angles
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: The answer is 8.
Step-by-step explanation: The first step is to convert the expression into figures. We shall call the unknown number Y. So if we are told “the square of a number,” that means Y squared, or better still, Y^2. Further we are told “the difference between the square of a number and 40” and that can be written as;
Y^2 - 40.
Next we are told that this expression is equal to 3 times that number (that is 3Y). That can now be written out as follows,
Y^2 - 40 = 3Y
If we move all expressions to one side of the equation, what we would have is,
Y^2 - 3Y -40 = 0
(Remember that when a positive value crosses to the other side of an equation it becomes negative and vice versa)
We now have a quadratic equation
Y^2 -3Y - 40 = 0
By factorizing we now have
(Y -8) (Y + 5) = 0
Therefore Y - 8 = 0 or
Y + 5 = 0
Hence, Y = 8 or Y = -5
Since we are asked to calculate the positive solution, Y = 8
The answer is 71 since you count from the tenth space. So the first significant figure would be 7 and the second significant figure would be 0. As we know that we round up one number of the previous number is 5 or bigger.
