No, the tangent line needs to be against the circle from where it leaves. This is just a straight up right angle.
A circle is a geometric object that has symmetry about the vertical and horizontal lines through its center. When the circle is a unit circle (of radius 1) centered on the origin of the x-y plane, points in the first quadrant can be reflected across the x- or y- axes (or both) to give points in the other quadrants.
That is, if the terminal ray of an angle intersects the unit circle in the first quadrant, the point of intersection reflected across the y-axis will give an angle whose measure is the original angle subtracted from the measure of a half-circle. Since the measure of a half-circle is π radians, the reflection of the angle π/6 radians will be the angle π-π/6 = 5π/6 radians.
Reflecting 1st-quadrant angles across the origin into the third quadrant adds π radians to their measure. Reflecting them across the x-axis into the 4th quadrant gives an angle whose measure is 2π radians minus the measure of the original angle.
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Oh wow- that is so funny... haha...
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Answer:
Jace~
Plus, here is a anime image that might make your day happy. . .
"Understand the problem" might rightly consist of
• Perform this step first
• Identify what you are being asked to solve or find.
• Identify the important words or numbers in the problem
• Identify any instructions that you are supposed to follow
_____
One of my professors always insisted we start the solution of any problem by writing down what was Given, and what we had to Find, using those headers for the sections of the paper we turned in. Only after those were listed were we allowed to write the Solution. Solution papers that didn't have that format were tossed in the trash, and no credit was given. Harsh, but effective.
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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