Help please with the question
1 answer:
You might be interested in
Place the decimal point right after the first nonzero digit, which is 3.
So you'll go from this
to this 
after this move.
Now ask yourself: "how many spots must I move the decimal point to get back to 0.003045?"
The answer is "3 spots to the left"
The exponent of the final answer is -3
The negative indicates "move left" and the 3 tells us how many places to move the decimal
Therefore,
Answer: Choice C)
This is assuming you meant to write the -3 as an exponent
So you'll go from this
Now ask yourself: "how many spots must I move the decimal point to get back to 0.003045?"
The answer is "3 spots to the left"
The exponent of the final answer is -3
The negative indicates "move left" and the 3 tells us how many places to move the decimal
Therefore,
Answer: Choice C)
This is assuming you meant to write the -3 as an exponent
Answer:
√3 is irrational
Step-by-step explanation:
The location of the third point of a triangle can be found using a rotation matrix to transform the coordinates of the given points.
<h3 /><h3>Location of point C</h3>With reference to the attached figure, the slope of line AC is √3, an irrational number. This means the line AC <em>never passes through a point with integer coordinates</em>. (Any point with integer coordinates would be on a line with rational slope.)
<h3>Equilateral triangle</h3>The line segments making up an equilateral triangle are separated by an angle of 60°. If two vertices are on grid squares, the third must be a rotation of one of them about the other through an angle of 60°. The rotation matrix is irrational, so the rotated point must have irrational coordinates.
The math of it is this. For rotation of (x, y) counterclockwise 60° about the origin, the transformation matrix is ...
Cos(60°) is rational, but sin(60°) is not. For any non-zero rational values of x and y, the sum ...
cos(60°)·x + sin(60°)·y
will be irrational.
As in the attached diagram, if one of the coordinates of the rotated point (B) is zero, then one of the coordinates of its image (C) will be rational. The other image point coordinate cannot be rational.
