a. The sum of the three numbers is 12, and A-1, B, C + 10 are equal ratio series Big brother, I can guess the numbers

a. The sum of the three numbers is 12, and A-1, B, C + 10 are equal ratio series Big brother, I can guess the numbers


2,4,6 or 17,4, - 9
b=4,a=8-c
∴(a-1)(18-a)=16



Let a


a/b=b/c
a+b+c=3
We can know 3B = 3, B = 1 by arithmetic sequence
According to the first formula, AC = 1
Then a + 1 / a = 2, a = 1, C = 1,
Also a



If the second, third and sixth items of the arithmetic sequence with tolerance not 0 constitute the arithmetic sequence, then the common ratio is______


a2*a6=a3^2
(a1+d)(a1+5d)=(a1+2d)^2
a1^2+6a1d+5d^2=a1^2+4a1d+4d^2
2a1d = - D ^ 2, tolerance is not 0
d=-2a1
q=a3/a2=(a1+2d)/(a1+d)
=(a1-4a1)/(a1-2a1)
=-3a1/(-a1)
=3



It is known that four real numbers - 9, A1, A2 and - 1 are equal difference sequences, and five real numbers - 9, B1, B2, B3 and - 1 are equal proportion sequences, then B2 (a2-a1) = ()
A. 8B. -8C. ±8D. 98


From the question, we get a & nbsp; 2 − A & nbsp; 1 = D = − 1 + 94 − 1 = 83, B22 = 9, and because B2 is the third term in the equal ratio sequence, it has the same sign with the first term, that is, B2 = - 3  B2 (a2-a1) = - 8



|If A-2 | and (a + 2b) &# 178; are opposite to each other, what is (b) &# 178; △ a equal to





How to measure the actual power of a 220 V 60 W incandescent lamp when it is connected to a home circuit


Turn off all the other appliances and watch the meter turn a few times



Rewrite 6.25 to the following number. What's the change of its size? 62.50.625 6250.0625


62.5 is 10 times of 6.25
625 is 0.1 times of 6.25
625 is 100 times of 6.25
0625 is 0.01 times of 6.25



Connect two resistors R1 = 15 ohm and R2 = 5 ohm in series to the power supply, and the voltage at both ends of R1 is 3V
The voltage at both ends of R2 is________ 5. The power supply voltage is______ V.


Because R1 and R2 are connected in series, there are:
UR1:UR2=R1:R2,
So UR2 = ur1xr2 / R1 = 1V,
The power supply voltage is ur1 + UR2 = 1 + 3 = 4V



A simple algorithm of 17 × 15:14 × 17:7 × 28:5


17×14/15 × 7/17 × 5/28
=(17× 7/17) ×(14/15×5/28)
=7×1/6=7/6



There is an ammeter with internal resistance Rg = 30 and full bias current Ig = 1mA. If you want to refit it into a voltmeter with a range of 0-3v, how much resistance should be connected in series? After refitting, it will be electrified


Full bias current Ig = 1mA, refit into a voltmeter to series a large resistance, how big, calculate. The maximum voltage divided by the full bias current is OK, r = 3V / 1mA = 3000 ohm, and because the original RG = 30, so as long as the series 3000-30 = 2970 ohm resistance is OK