answer:To find the value of X, we need to solve the equation: 1.5 * [ ( X * 0.48 * 2.50 ) / ( 0.12 * 0.09 * 0.5 ) ] = 1200.0000000000002 First, let's simplify the denominator: 0.12 * 0.09 * 0.5 = 0.0054 Now, the equation becomes: 1.5 * [ ( X * 0.48 * 2.50 ) / 0.0054 ] = 1200.0000000000002 Next, let's simplify the numerator inside the brackets: X * 0.48 * 2.50 = 1.2X So the equation is now: 1.5 * ( 1.2X / 0.0054 ) = 1200.0000000000002 Now, let's solve for X: 1.5 * ( 1.2X / 0.0054 ) = 1200.0000000000002 ( 1.8X / 0.0054 ) = 1200.0000000000002 1.8X = 1200.0000000000002 * 0.0054 1.8X = 6.480000000000001 Now, divide both sides by 1.8 to solve for X: X = 6.480000000000001 / 1.8 X = 3.6000000000000005 Therefore, the value of X is approximately boxed{3.6} .

answer:1. Given that ( f, g, varphi: mathbb{R} to mathbb{R} ) are ascendant functions, we know that for any ( x, y in mathbb{R} ) with ( x leq y ), it holds that ( f(x) leq f(y) ), ( g(x) leq g(y) ), and ( varphi(x) leq varphi(y) ). 2. We are also given that ( f(x) leq g(x) leq varphi(x) ) for all ( x in mathbb{R} ). 3. To prove ( f(f(x)) leq g(g(x)) leq varphi(varphi(x)) ) for all ( x in mathbb{R} ), we start by considering the first part of the inequality ( f(f(x)) leq g(g(x)) ). 4. Since ( f ) is ascendant and ( f(x) leq g(x) ), applying ( f ) to both sides of the inequality ( f(x) leq g(x) ) gives: [ f(f(x)) leq f(g(x)) ] 5. Next, since ( f(x) leq g(x) ) and ( g ) is ascendant, applying ( g ) to both sides of the inequality ( f(x) leq g(x) ) gives: [ g(f(x)) leq g(g(x)) ] 6. However, we need to show ( f(f(x)) leq g(g(x)) ). Notice that from step 4, we have ( f(f(x)) leq f(g(x)) ), and from step 5, we have ( f(g(x)) leq g(g(x)) ). Combining these two results, we get: [ f(f(x)) leq f(g(x)) leq g(g(x)) ] 7. Therefore, we have shown that ( f(f(x)) leq g(g(x)) ). 8. Now, consider the second part of the inequality ( g(g(x)) leq varphi(varphi(x)) ). 9. Since ( g(x) leq varphi(x) ) and ( g ) is ascendant, applying ( g ) to both sides of the inequality ( g(x) leq varphi(x) ) gives: [ g(g(x)) leq g(varphi(x)) ] 10. Next, since ( g(x) leq varphi(x) ) and ( varphi ) is ascendant, applying ( varphi ) to both sides of the inequality ( g(x) leq varphi(x) ) gives: [ varphi(g(x)) leq varphi(varphi(x)) ] 11. However, we need to show ( g(g(x)) leq varphi(varphi(x)) ). Notice that from step 9, we have ( g(g(x)) leq g(varphi(x)) ), and from step 10, we have ( g(varphi(x)) leq varphi(varphi(x)) ). Combining these two results, we get: [ g(g(x)) leq g(varphi(x)) leq varphi(varphi(x)) ] 12. Therefore, we have shown that ( g(g(x)) leq varphi(varphi(x)) ). 13. Combining the results from steps 7 and 12, we have: [ f(f(x)) leq g(g(x)) leq varphi(varphi(x)) quad forall x in mathbb{R} ] (blacksquare)

answer:Firstly, we factor the given quadratic inequality: [ -y^2 + 9y - 20 = -(y^2 - 9y + 20) ] Factoring the quadratic expression inside the parenthesis: [ y^2 - 9y + 20 = (y - 4)(y - 5) ] So the inequality becomes: [ -(y - 4)(y - 5) < 0 ] This implies: [ (y - 4)(y - 5) > 0 ] Analyzing this, we see: - If ( y < 4 ), then both ( y - 4 ) and ( y - 5 ) are negative, so their product is positive. - If ( 4 < y < 5 ), ( y - 4 ) is positive and ( y - 5 ) is negative, thus their product is negative. - If ( y > 5 ), both ( y - 4 ) and ( y - 5 ) are positive, so their product is positive. Using this analysis, the inequality ( (y - 4)(y - 5) > 0 ) is satisfied for ( y < 4 ) or ( y > 5 ). Hence, the interval notation for the solution is: [ boxed{(-infty, 4) cup (5, infty)} ]

answer:If the pig is only half the size of an average pig, then the farmer will get half the amount of bacon from it. So instead of 20 pounds, he will get 10 pounds of bacon from the runt pig. The farmer makes 60 from the 10 pounds of bacon. To find out how much he sells each pound for, we divide the total amount made by the number of pounds: 60 / 10 pounds = 6 per pound The farmer sells each pound of bacon for boxed{6} .