Answer:
an element with a small number of valence electrons and a large atomic radius
Answer:
The law of multiple proportions states that when two elements can combine in different ratios to form different compounds, the masses of the element combining with the fixed mass of another element result in whole number ratios. This shows that the law of multiple proportions is followed
Explanation:
It's the natural tendency of things to keep going unless there's something trying to stop them.
It's usually called "inertia".
Don't get the idea from all of this that things stop unless there's something to keep them going. The truth is exactly the opposite: Things keep going unless there's something to make them stop.
For the answer to the question above,
I'll write down my solution to your problem
sin(A+B) = sinAcosB + cosAsinB
<span>sin(2A) = 2sinAcosA </span>
<span>cos(2A) = 1-2sin^2A </span>
<span>sin(3x) = sin(2x+x) </span>
<span>sin(3x) = sin(2x)cos(x) + cos(2x)sin(x) </span>
<span>= 2sin(x)cos(x)cos(x) + (1-2sin^2(x))sin(x) </span>
<span>= 2sin(x)cos^2(x) + sin(x) - 2sin^3(x) </span>
<span>= 2sin(x)(1-sin^2(x)) + sin(x) - 2sin^3(x) </span>
<span>= 2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x) </span>
<span>= 3sin(x) - 4sin^3(x)
</span>My closest answer is multiple choice letter D.
The complete observation about adding bulb 3 is the brightness of the bulbs has to do with power which considers both the voltage and the current: less voltage x less current = dimmer bulbs. In circuit A, the voltage is divided across the resistors and the current decreases as resistance increases. In circuit B, the voltage is the same in each parallel section of the circuit and the current through that section of the circuit only depends on the resistor in that section.
<h3>What is power of the circuit?</h3>
The power of the bulb or any resistor is equal to the product of voltage and current flowing through it.
P = VI
Circuit A has bulbs in series while the circuit B has bulbs in parallel.
When bulb 3 added to circuit A, the brightness of all the bulbs dimmed but when bulb 3 (R3) added to circuit B, nothing changed in the brightness of the bulb.
The brightness is depended on the power of the circuit. When both the voltage and current are less, the bulb will be dimmed. In circuit A, series resistors divide the voltage across them. In circuit B, voltage is equal for all the resistors.
Thus, the last option is correct.
Learn more about power.
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