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Basile [38]
3 years ago
10

P=I squared R solve for I

Mathematics
1 answer:
jeka943 years ago
5 0

l^2R=P\ \ \ \ |:R\neq0\\\\l^2=\dfrac{P}{R}\to l=\pm\sqrt{\dfrac{P}{R}}

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Solve the following equation X = 2 and y = 4 <br><br> 3x + 2y + 4 =?
svetoff [14.1K]

Answer:

18

Step-by-step explanation:

3(2)+2(4)+4=

6+8+4=18

6 0
3 years ago
Read 2 more answers
Consider a sphere of radius R &gt; 0. Take two antipodal points A and B on the sphere, and another point C on the sphere. Check
aleksklad [387]

Answer:

To this question; we want to show that if two antipodal points on a sphere were A and B, for any random point C on the sphere, AC is perpendicular to BC?

Step-by-step explanation:

I would proceed by imagining the sphere in a three dimensional Cartesian coordinate system. For example, use a sphere of diameter 2 and let it sit at the origin. Then (0, 0, 1) and (0, 0, -1) are the vector locations of the “north and south poles” of the sphere.

Now choose the vector location of any point on the surface of the sphere. It will have a vector location - we’ll call it (x, y, z). Now the vectors from your point to the two poles are (-x, -y, 1-z) and (-x, -y, -1-z).

Now just form the dot product of those two vectors:

(-x, -y, 1-z) . (-x, -y, -1-z) = x^2 +y^2 + (1-z)*(-1-z)

Now the truth of your claim will be embodied in that dot product being zero:

x^2 + y^2 - (1+z)(1-z) = 0

x^2 + y^2 - (1-z^2) = 0

x^2 + y^2 + z^2 = 1

But that last line is just the definition of points on the surface of a sphere of radius 1, so the claim is proven.

Since R>0 and AC is perpendicular to BC, the <ACB is at right angle.

Please, find attached a simple image to show antipodal points (Two points that makes a diameter) and a point C on the sphere.

5 0
3 years ago
Please help me the photo is up above
wolverine [178]

Answer:

The Correct Symplified ratio is 1:2

Step-by-step explanation:

5 0
2 years ago
What does i^16 equal and how do you get the answer​
svp [43]

Answer:

i^16 = 1

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i = √-1 = i

i^2 = i * i = (√-1)^2 = -1

i^3 = i^2 * i^1 = -1 * √-1 = -√-1 or -i

i^4 = i^2 * i^2 = -1 * -1 = 1

the order keeps following......

i^16 = 1

since i^16 is same thing as i^4 * i^4

and we know i^4 = 1 so 1*1=1

7 0
3 years ago
Can you think of a situation where comparing two fractions directly is easier than converting them to decimal or percents in ord
Andrei [34K]

when you are comparing 1/2 to 3/6. this is because you know that 1/2 is equal to 3/6. You are very welcome, and thank you for the challenge.

6 0
3 years ago
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