answer:Since point P (m+1, m+4) is on the y-axis of the Cartesian coordinate system, it follows that m+1=0. Solving this, we get m=-1, Therefore, the coordinates of point P are (0, 3). Hence, the answer is boxed{(0, 3)}.

answer:1. **Understanding the problem**: Jasmine originally had enough paint for 50 rooms but losing 4 cans decreased her coverage to 36 rooms. The additional cans allow her to cover 2 rooms per can. 2. **Calculating the lost capacity per original can**: [ text{Loss in room coverage} = 50 - 36 = 14 text{ rooms} ] [ text{Number of lost original cans} = 4 ] From this, each of the original lost cans would have covered [ text{Rooms per lost can} = frac{14}{4} = 3.5 ] 3. **Calculating how many additional cans were needed**: Jasmine initially could cover 36 rooms. To cover all 50 rooms, she still needed to gain coverage for [ 50 - 36 = 14 text{ rooms} ] With each of the second type of can covering 2 rooms, [ text{Number of additional type cans} = frac{14}{2} = 7 ] 4. **Conclusion (with the boxed answer)**: Jasmine used 7 cans of the second type to make up for the shortfall and cover all 50 rooms. The final answer is boxed{C. 7 cans}

answer:To find out how many books John sold on Thursday, we first need to determine the total number of books that were not sold, which is 63.45% of the initial stock. The initial stock of books is 1100. The percentage of books not sold is 63.45%. To find the number of books not sold, we calculate: Number of books not sold = (63.45/100) * 1100 Now, let's calculate the number of books not sold: Number of books not sold = 0.6345 * 1100 Number of books not sold = 697.95 Since we cannot have a fraction of a book, we'll round this to the nearest whole number. However, since the percentage given should be exact, we'll keep it as 698 books not sold for the purpose of this calculation. Now, let's find out how many books were sold in total from Monday to Friday. We subtract the number of books not sold from the initial stock: Total books sold = Initial stock - Number of books not sold Total books sold = 1100 - 698 Total books sold = 402 We know the number of books sold from Monday to Wednesday and on Friday. Let's add those up: Books sold from Monday to Wednesday and Friday = 75 (Monday) + 50 (Tuesday) + 64 (Wednesday) + 135 (Friday) Books sold from Monday to Wednesday and Friday = 324 Now, we subtract the number of books sold from Monday to Wednesday and on Friday from the total number of books sold to find out how many were sold on Thursday: Books sold on Thursday = Total books sold - Books sold from Monday to Wednesday and Friday Books sold on Thursday = 402 - 324 Books sold on Thursday = 78 Therefore, John sold boxed{78} books on Thursday.

answer:(I) The inequality f(x) geqslant g(x) holds true for all x in R, which can be rewritten as (x^{2}-1) geqslant a|x-1|(*). quadquad ① When x=1, (*) is obviously true, and a in R. quadquad ② When x neq 1, (*) can be transformed into a leqslant frac{x^{2}-1}{|x-1|}. Let varphi(x)=frac{x^{2}-1}{|x-1|}= begin{cases} x+1,(x > 1) -(x+1),(x < 1) end{cases}. Since varphi(x) > -2 for all x, we have a leqslant -2 in this case. Combining ① and ②, the range of values for the real number a is a leqslant -2. (II) h(x)=|f(x)|+g(x)=|x^{2}-1|+a|x-1|= begin{cases} x^{2}+ax-a-1,(xgeqslant 1) -x^{2}-ax+a+1,(-1leqslant x < 1) x^{2}-ax+a-1,(x < -1) end{cases} We will now consider different cases to find the maximum value of h(x) on the interval [-2,2]: quadquad ① When a > 2, h(x) is decreasing on [-2,1] and increasing on [1,2]. The maximum value in this case is 3a+3. quadquad ② When 0 leqslant a leqslant 2, h(x) is decreasing on [-2,-1] and [- frac{a}{2},1], and increasing on [-1,- frac{a}{2}] and [1,2]. The maximum value in this case is 3a+3. quadquad ③ When -2 leqslant a < 0, h(x) is decreasing on [-2,-1] and [- frac{a}{2},1], and increasing on [-1,- frac{a}{2}] and [1,2]. The maximum value in this case is a+3. quadquad ④ When -3 leqslant a < -2, h(x) is decreasing on [-2, frac{a}{2}] and [1,- frac{a}{2}], and increasing on [ frac{a}{2},1] and [- frac{a}{2},2]. The maximum value in this case is a+3. quadquad ⑤ When -4 < a < -3, h(x) is decreasing on [-2, frac{a}{2}] and [1,- frac{a}{2}], and increasing on [- frac{a}{2},1] and [- frac{a}{2},2]. The maximum value in this case is 0. quadquad ⑥ When a leqslant -4, h(x) is increasing on [-2,1] and decreasing on [1,2]. The maximum value in this case is 0. In summary, the maximum value of h(x) on [-2,2] is: boxed{begin{cases} 3a+3, (a geqslant 0) a+3, (-3 leqslant a < 0) 0, (a < -3) end{cases}}