Is 9 out of 10 close to 1 or 1 out of 2

Is 9 out of 10 close to 1 or 1 out of 2

Hello, the answer is 1
If a5-a1 = 15, a4-a2 = 6, then A3 = ()
A. 4B. -4C. ±4D. ±2
Let the common ratio of the equal ratio sequence be q, then ∵ a5-a1 = 15, a4-a2 = 6, ∵ a1q4-a1 = 15, a1q3-a1q = 6, ∵ Q2 + 1 = 52q ∵ q = 2 or q = 12 ∵ A1 = 1 or A1 = - 16 ∵ A3 = ± 4, so C is selected
The distance between a and B is 90 kilometers, and the speed of the car is twice that of the truck. The two cars start from place a at the same time, and the car returns immediately after it arrives at place B,
(following the above question) then the two cars meet at C. how many kilometers is the distance between B and C?
The two cars travel 90 × 2 = 180 km
The distance of a small car is twice that of a big car
So the small car runs 180 × 2 ^ (2 + 1) = 120 km
120-90=30
A: the distance between B and C is 30 kilometers
Speed ratio of car to truck = 2:1
Then the distance ratio is 2:1
When the two cars meet, they travel a total of 90 × 2 = 180 km, which is the whole journey
Then a line 180 × 1 / (2 + 1) = 60 km
Distance between B and C = 90-60 = 30 km
The two cars traveled a total of 180 km, the car 120 km, the truck 60 km, and the distance between C and B was 30 km.
The two cars travel 90 × 2 = 180 km
The distance of a small car is twice that of a big car
So the small car runs 180 × 2 ^ (2 + 1) = 120 km
120-90=30
Calculate (1 + 1.2) + (2 + 1.2 * 2) + (3 + 1.2 * 3) ((100 + 1.2 * 100) with simple method=
（1+1.2）+（2+1.2*2）+（3+1.2*3）.（100+1.2*100）
=(1+2+3+...+100)+(1.2+1.2*2+...+1.2*100)
=(1+2+3+...+100)+1.2*(1+2+...+100)
=(1+2+3+...+100)*(1+1.2)
=(1+100)*100/2*2.2
=11110
1, 0.5 what should be filled in the air
Let an = an ^ 2 + BN + C
2=a+b+c.
1=4a+2b+c,
0.5=16a+4b+c,
Subtraction gives - 1 = 3A + B,
-0.5=12a+2b,
The solution is a = 1 / 4, B = - 7 / 4, C = 7 / 2
∴an=(1/4)(n^2-7n+14),
a3=0.5.
2,1,0.5, 0.5 can be filled in the air
Because A1 = 2 / 1 = 2, A2 = 2 / 2 = 1, A3 = 2 / 3, A4 = 2 / 4 = 1 / 2 = 0.5, so an = 2 / n.
2,1,0.5, 0.5 can be filled in the air
okay
Just the sauce
Fill in 2 / 3
The rule is that the nth term equals 2 divided by N, so the 3rd term equals 2 divided by 3
The speed of the car is 80 km / h, the speed of the truck is 65 km / h, when the two cars meet
The car is more than two kilometers bigger than the truck
Yu Ting 15,
Let two cars meet in X hours
80X－45＝65X＋45
80X－65X＝45×2
15X＝90
X＝6
Simple calculation of 11 / 13 / 8 + 3 / 4 / 8
11 / 13 ÷ [(5 / 8 + 3 / 4) / / 5 / 8]
11 / 13 ÷ [(5 / 8 + 3 / 4) / / 5 / 8]
=11/13÷（5/8÷5/8+3/4÷5/8）
=11/13÷（1+6/5）
=11/13÷11/5
=5/13
Finding regular sequence
18 10 6 4 （）
18 10 6 4 （3）
Divide 18 by 2 to get 9; add 1 to get 10
Divide 10 by 2 to get 5; add 1 to get 6
and so on
Both small cars and large cars have to drive from place a to place B. the speed of large cars is 80% of that of small cars, and large cars have to stop at the middle point of two places for 10 minutes
Both small cars and large cars have to drive from place a to place B. the speed of large cars is 80% of that of small cars. Large cars have to stop at the middle point of two places for 10 minutes. Small cars don't stop halfway, but they start 11 minutes later than big cars, but they arrive at place B 7 minutes earlier than big cars. It is known that big cars start at 10 o'clock in the morning, Thank you
It took X minutes to set up a car trip
（x-10+7+11)*80%=x*100%
0.8x+6.4=x
0.2x=6.4
x=32
32+11=43
Then the car will arrive at B at 10:43
Three mathematical problems
9*1.7+9.1/1.7-5*1.7+4.5/1.7 3.7*15+21*4.5 98989898*99999999/1010101/11111111
9×1.7+9.1÷1.7-5×1.7+4.5÷1.7=(9-5)×1.7+(9.1+4.5)÷1.7=4×1.7+13.6÷1.7=6.8+8=14.83.7×15+21×4.5=3.7×15+6.3×15=10×15=15098989898×99999999÷1010101÷11111111=(98989898÷1010101)×(99999999÷111...
1. (9-5)×1.7+（9.1+4.5）÷1.7=6.8+8=14.8
2. 3.7×15+6.3×15=15×（3.7+6.3）=150
3. （98989898÷ 1010101）×（99999999÷11111111）
=98×9=882
9×1.7+9.1÷1.7-5×1.7+4.5÷1.7
=(9-5)×1.7+(9.1+4.5)÷1.7=4×1.7+13.6÷1.7=14.8
3.7×15+21×4.5=3.7×15+6.3×15=150
98989898×99999999÷1010101÷11111111=(98989898÷1010101)×(99999999÷11111111=882