The new definition operation stipulates that if a * k = ax (a + 1) + (a + 2) +... + (a + 1), then (8 * 7) * 3 is equal to

The new definition operation stipulates that if a * k = ax (a + 1) + (a + 2) +... + (a + 1), then (8 * 7) * 3 is equal to

The title is not clear, a +... Does this part last to k?
8*7=8x（8+1）+（8+2）+...(8+7)+...(8+1)=216
Polynomial = 216 * 3 = 216x (216 + 1) + (216 + 2) + (216 + 3) + (216 + 2) + (216 + 1) = 47744
Is LZ asking such a question? Should I have misunderstood it?

If the voltage at both ends of a fixed value resistor increases from 8V to 10V and the current through the resistor changes by 0.1A, the electric power of the resistor changes____
I know this problem has been raised, but the question is why can't we directly multiply the voltage change by the current change? The point is, please explain this problem clearly

Assuming that the original voltage U1, the original current is I1; when the voltage rises to U2, the current changes to I2, the change of electric power is as follows:
△P=P2-P1=U2²/R-U1²/R=（U2²-U1²）/R=（U2-U1）（U2+U1）/R=△U（U2+U1）/R.
Change of current: △ I = i2-i1 = U2 / r-u1 / r = (u2-u1) / r = △ U / R
The current change multiplied by the voltage change is: △ u ×△ I = (△ U) &
∵U2+U1≠△U
∴△P≠△U×△I

Calculate M-N + [(2n & # 178;) / (M + n)]

Primitive = [(m-n) (M + n) + 2n & # 178;] / (M + n)
=(m²-n²+2n²)/(m+n)
=(m²+n²)/(m+n)

The maximum value of a sinusoidal AC voltage is 311V. If it is used to supply power to the consumer with 110 ohm resistance, the voltage in parallel with the ammeter connected in series with the consumer in the circuit will be reduced
The indication is correct
A voltmeter 311V
B voltmeter 220 V
C ammeter 2.82a
D ammeter 1.41a
Still hope to analyze

The data measured by Voltmeter and ammeter are valid values
So the voltmeter is 220 v
The ammeter is 220 / 110 = 2A

Given that the line y = kx-2 and the parabola y square = 8x intersect at two points a and B, if the abscissa of the midpoint of the line AB is 2, what is ab equal to
Now the exam, everyone help. Help me, I will reward you with the computer

Let a (x1, Y1) B (X2, Y2) get: (x1 + x2) / 2 = 2 = = > (x1 + x2) = 4 simultaneous equations: y = kx-2; y ^ 2 = 8x, get: K ^ 2x ^ 2 - (4K + 8) x + 4 = 0. According to Weida's theorem: X1 + x2 = (4K + 8) / (k ^ 2), x1x2 = 4 / (k ^ 2), and X1 + x2 = 4, get k = - 1 or K = 2, so x1x2 = 4 / (k ^ 2) = 4 or 1

The resistances of the two wires are 10 ohm and 5 ohm respectively. If they are connected in series to a power supply, the ratio of heat generated by current passing through the two wires is -------- in the same time. If they are connected in parallel to a power supply, the ratio of heat generated by current passing through the two wires is -------- in the same time?

The series current is equal, the heat is the ratio of the resistance, q = I square RT. the parallel voltage is equal, the heat is the ratio of the reciprocal of the resistance, q = u square T / R

(3/2)∫ 1/[(x-1/2)²+3/4] dx=?

How much Joule energy does a 500W electric rice cooker consume per minute when it is used under rated voltage

500*60=30000J

In the equal ratio sequence, all are positive. A3 * A9 = 4, A6 * A10 + a3 * A5 = 41. Find A4 + A8

a3*a9=a4*a8=4
a6*a10=a8²
a3*a5=a4²
therefore
a4*a8=4
a8²+a4²=41
a4+a8>0
(a4+a8)²
=a4²+2a4*a8+a8²
=41+2*4
=49
a4+a8=7

If two electric bulbs of "220 V, 100 W" and "110 V, 100 W" are connected in series in a circuit, the voltage ratio at both ends of the bulb is 0

R1=4R2
U1=4U2
U1/U2=4/1