C.) is the best answer others are less then 50%.
C.) $35 is $15 more then 20 this is over 50%
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Answer:
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Step-by-step explanation:
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Step-by-step explanation:
<em>Let </em><em>the </em><em>two </em><em>numbers </em><em>be </em><em>x </em><em>and </em><em>y </em>
<em>x </em><em>-</em><em> </em><em>y </em><em>=</em><em> </em><em>6</em><em>8</em><em>5</em>
<em>Let </em><em>the </em><em>smaller </em><em>number </em><em>be </em><em>y </em>
<em>x </em><em>-</em><em> </em><em>2</em><em>6</em><em>2</em><em> </em><em>=</em><em> </em><em>6</em><em>8</em><em>5</em>
<em>x </em><em>=</em><em> </em><em>6</em><em>8</em><em>5</em><em> </em><em>+</em><em> </em><em>2</em><em>6</em><em>2</em>
<em>Therefore </em><em>x </em><em>=</em><em> </em><em>9</em><em>4</em><em>7</em>
Answer:
a. 
b. 
c. 
Step-by-step explanation:
Given:
,
and 



a.
×
A
b.
×
A
c.
A
The LED which would glow brightest is LED C with the greatest current and voltage
The LED which would be the most dim is LED B with low voltage and consequently low current.
<h3>How to determine the current</h3>
The formula for finding current
I = V/R
Where v = voltage
R = resistance
A. V = 12V
R = 4. 7 + 15 = 19. 7 kΩ = 19700 Ω in series

×
A
B. V = 9V
R = 4. 7 + 1 = 4. 7 kΩ = 4700Ω in series

×
A
C. V= 12V
1/R =
=
× 
R =
= 310. 56 Ω

A
It is important to note that the brightness of a bulb depends on both current and voltage depending on whether the bulb it is in parallel or series.
The LED which would glow brightest is LED C with the greatest current and voltage
The LED which would be the most dim is LED B with low voltage and consequently low current.
Learn more about Ohms law here:
brainly.com/question/14296509
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