(10*10+11*11+12*12+13*13+14*14)/365=

jien11182022-10-04 11:39:543条回答

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hehanyong21 共回答了18个问题 | 采纳率83.3%
(10*10+11*11+12*12+13*13+14*14)/365
=【1/6*14(14+1)*(14*2+1)-1/6*9*(9+1)*(9*2+1)】/365
=【1/6*14*15*29-1/6*9*10*19】/365
=【1015-285】/365
=730/365
=2
1年前
kkkk0512 共回答了692个问题 | 采纳率
(10*10+11*11+12*12+13*13+14*14)/365
=(100+121+144+169+196)/365
=730/365
=2
1年前
秋风瑟瑟秋雨绵绵 共回答了49个问题 | 采纳率
2
1年前

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(1)15.3+0.86+0.14+4.7 (2)365×99 (3)125×64 (4)0.25×44 (5)19.6
(1)15.3+0.86+0.14+4.7 (2)365×99 (3)125×64
(4)0.25×44 (5)19.6÷2÷5 (6)2.73×4×2.5
(7)13.5+(15-9.75) (8)(800-480÷12)×25 (9)176-76÷4+66
回到原处1年前1
fairyjin 共回答了24个问题 | 采纳率91.7%
(1)原式=15.3+4.7+(0.86+0.14)
=20+1
=21 (2)原式 =365×(100-1)
=36500-365
=36135 (3)原式=125×8×8
=8000

(4)原式=0.25×4×11
=11 (5)原式=19.6÷(2×5)
=1.96 (6)原式=4×2.5×2.73
=27.3
(7)原式=13.5+15-10+0.25
=13.5+5+0.25
=18.75 (8)原式=(800-40)×25
=20000-1000
=19000 (9)原式=176-19+66
=176-20+1+66
=223
I’ve spent over a year in India, and in those 365 plus days,
I’ve spent over a year in India, and in those 365 plus days, I’ve learned a lot about getting around Indian cities. My biggest lessons have been learned through being cheated, particularly by taxi and rickshaw (人力车) drivers, but that doesn’t mean those are bad ways to travel, as long as you know what you’re doing. Below are the best ways to get around the city of Delhi, India, and tips for how to keep from being the victim of scams (欺骗).
Taking taxis is a great way to get around the city of Delhi and chances are, if you arrive in Delhi by plane, as soon as you make it through customs, you’ll be swarmed by Indian taxi drivers. At the Delhi airport, be sure to arrange for a taxi to your hotel at one of the two Delhi Traffic Police Taxi Booths. One is inside the airport, and one is outside. The key is to make sure to go to a booth run by the police, rather than by independent taxi drivers.
Rickshaws are one of my favorite ways to get around Indian cities, partly because it’s how the locals often travel. Auto-rickshaws are more common, but bicycle rickshaws are still used in Old Delhi. If you do have a chance to take a bicycle rickshaw, you should do it at least once for a unique experience that should only set you back about 15 rupees. Auto-rickshaw rates around Delhi range between 30 and 80 rupees, depending on distance.
If you really want to travel around Delhi like the locals, take a public bus. Indian buses become very crowded and most do not have air conditioning. They are, however, very cheap. A bus trip won’t set you back any more than 15 rupees, as long as you stay within the city limits. Since Indian buses get so crowded, try to board the bus at the start of the route so you can get a seat.
The train is a great way to get around within the city of Delhi. Fares are reasonable, between six and 22 rupees. All departure announcements are in both Hindi and English, and tokens can be purchased for between six and 22 rupees.
小题1: What is the author trying to do through this text?
A.Expect us to travel around Delhi.
B.Show his/her experiences in Delhi.
C.Give some advice of traveling in Delhi.
D.Explain the difficulties of traveling in Delhi.
小题2:What should you do to avoid being cheated when taking a taxi at the Delhi airport?
A.Go to a police-run booth.
B.Go out of the airport.
C.Show your ticket to the driver.
D.Pay more to the drivers to keep safe.
小题3:The author suggests taking a rickshaw in order to ______.
A.save some money
B.gain some unique experience
C.enjoy the comfortable trip
D.help the local rickshaw drivers
小题4:Which may be the topic that follows?
A.Car rentals in Delhi
B.Food and drink in Delhi
C.Weather conditions in Delhi
D.Hotel recommendations in Delhi
congcong111年前1
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小题1:C
小题2:A
小题3:B
小题4:A

365+365+365+365+366=
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1461.
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=365×(0.64+0.35+0.01)
=365×1
=365
c=((year-1)*365+((year-1)/4-(year-1)/100+(year-1)/400+1))%7;
c=((year-1)*365+((year-1)/4-(year-1)/100+(year-1)/400+1))%7;这是什么意思?
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公元元年(也就是第一年)的第一天是星期1,以后的每一年与元年的差值取模7就可以算出该年的第一天是星期几.
c=[365*(year-1)+其中闰年的个数(闰年多一天)]%7+1;
c=((year-1)*365+((year-1)/4-(year-1)/100+(year-1)/400+1))%7;其中((year-1)/4-(year-1)/100+(year-1)/400就是其中闰年的个数,四年一闰,百年不闰,四百年再闰,所以4年的个数减去100年的个数在加上400年的个数就是其中闰年的个数了;
因为365=364+1;364%7=0;且后面的1可以加到去模公式前面去;所以可以化简成
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(2x+5)×50+y-365=1891 求XY
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361+362+363+364+365+366+367+368+369=365*
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361+369=365x2
362+368=365x2
363+367=365x2
364+366=365x2
综上 361+362+363+364+365+366+367+368+369=365x9
1095*(x/10)+365 * 10 * 1 * 0.4 ≤ 1095 * (1+30%)+365 * 10 * 0
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1095*(x/10)+365 * 10 * 1 * 0.4 ≤ 1095 * (1+30%)+365 * 10 * 0.55*0.4
109.5x+1460 ≤ 1423.5+803
109.5x ≤2226.5-1460
109.5x ≤766.5
x ≤766.5÷109.5
x ≤7
10*10+11*11+12*12+13*13+14*14/365=?
10*10+11*11+12*12+13*13+14*14/365=?
如何进行口答?
jiangtaowu1年前8
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你写的算式有问题,应该是:
(10*10+11*11+12*12+13*13+14*14)/365 =2
这道题不是胡乱来的.我是在新编十万个为什么上看到的这道题.这个是个数学家做普及教育的时候给小朋友出的题.
不过,你们老师要求口答简直是无理要求.
这道题实际上是有背景的:
你可以观察到:
10*10+11*11+12*12=13*13+14*14 = 365
这个式子有理论背景的.
类似的等式也可以写出一堆.
不过七年级的话,证明起来也没用.
实际上讲,只要明白,
只需算出10*10 + 11*11 +12*12=365,结果就是365/365*2 = 2就可以了
另外:
还有个公式:
1*1+2*2+..+n*n = n*(n+1)*(2n+1)/6
你可以用n=14,得到1*1+...+14*14=14*15*29/6
然后n=9,得到1×1+...+9*9=9*10*19/6
两者相减得到分子.不过方法肯定没有上面的简单
211×(-455)+365×455-211×545+545×365=______.
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羽丫头 共回答了15个问题 | 采纳率86.7%
解题思路:首先分解因式变为455×(-211+365)+545×(-211+365),然后再分解因式即可求解.

211×(-455)+365×455-211×545+545×365
=455×(-211+365)+545×(-211+365)
=(-211+365)(455+545)
=154×1000
=154000.
故答案为:154000.

点评:
本题考点: 有理数的混合运算.

考点点评: 此题主要考查了有理数的混合运算,解题的关键是利用因式分解简化计算,使计算变得简单.

521+365=886 224呢
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原句翻译过来是我爱你365天等于跟你说拜拜,
那么意思就是:我爱你三百百六十五天是不可能的
70乘24乘365乘60
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365x=365,x=?
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361+362+363+364+365+366+367
361+362+363+364+365+366+367
=( )乘( ) =( )
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361+362+363+364+365+366+367
=(364 )乘(7 ) =( 2548)
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13×4+x=365!
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原方程可变为:
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361+362+363+364+365+366+367=( )×( )=( )
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这七个数是连着的,他们的平均数就是中间的364
所以结果就是364*7=2548
4+5+6+7+8+…+80 9+19+29+39+49+59+69 465+476-365. 11+12+13+…+2
4+5+6+7+8+…+80 9+19+29+39+49+59+69 465+476-365. 11+12+13+…+20 987-491-187.
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1*5=4 2*6=7 51*69=365 5*1=?
1*5=4 2*6=7 51*69=365 5*1=?
不会啊
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1*5=4 5*1当然=5
=PMT(D15/12*365/360,D18,-D6),
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PMT 函数
基于固定利率及等额分期付款方式,返回贷款的每期付款额.
语法
PMT(rate,nper,pv,fv,type)
Rate 贷款利率.
Nper 该项贷款的付款总数.
Pv 现值,或一系列未来付款的当前值的累积和,也称为本金.
Fv 为未来值,或在最后一次付款后希望得到的现金余额,如果省略 fv,则假设其值为零,也就是一笔贷款的未来值为零.
Type 数字 0 或 1,用以指定各期的付款时间是在期初还是期末.
Type 值 支付时间
0 或省略 期末
1 期初
说明
PMT 返回的支付款项包括本金和利息,但不包括税款、保留支付或某些与贷款有关的费用.
应确认所指定的 rate 和 nper 单位的一致性.例如,同样是四年期年利率为 12% 的贷款,如果按月支付,rate 应为 12%/12,nper 应为 4*12;如果按年支付,rate 应为 12%,nper 为 4.
提示
如果要计算贷款期间的支付总额,请用 PMT 返回值乘以 nper.
211×(-455)+365×455-211×545+545×365=______.
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解题思路:首先分解因式变为455×(-211+365)+545×(-211+365),然后再分解因式即可求解.

211×(-455)+365×455-211×545+545×365
=455×(-211+365)+545×(-211+365)
=(-211+365)(455+545)
=154×1000
=154000.
故答案为:154000.

点评:
本题考点: 有理数的混合运算.

考点点评: 此题主要考查了有理数的混合运算,解题的关键是利用因式分解简化计算,使计算变得简单.

0.365*640+3又1/3*36.5+365*0.01
0.365*640+3又1/3*36.5+365*0.01
是3又3分之1,不准改
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0.365×640+10/3×36.5+365×0.01=0.365×(640+1000/3+10)=0.365×2950/3=1076.75/3=4307/12
211*(-455)+365*455-211*545+545*365
211*(-455)+365*455-211*545+545*365
sorry 刚刚忘记了
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=(365-211)*455+545*(365-211)
=154*455+154*545
=154*1000
=154000
211×(-455)+365×455-211×545+545×365=______.
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解题思路:首先分解因式变为455×(-211+365)+545×(-211+365),然后再分解因式即可求解.

211×(-455)+365×455-211×545+545×365
=455×(-211+365)+545×(-211+365)
=(-211+365)(455+545)
=154×1000
=154000.
故答案为:154000.

点评:
本题考点: 有理数的混合运算.

考点点评: 此题主要考查了有理数的混合运算,解题的关键是利用因式分解简化计算,使计算变得简单.

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