The number of positive 3-digit numbers have a property that the first digit is at least three times the second digit is 285.
<h3>
How to get the number?</h3>
Since restriction is set to the first digit, all you have to think is about the last two digits e.g if the first digit is one, the last two digits can only be zeros.
Thus, by the principle of multiplication, 1×1=1 way. If the first digit is 2, then the last digits can be either 1 or 0. You have 2×2=4 ways
For first digit 3, the last two digits can be 0,1 or 2. 3x3=9, there is a succinct pattern. Therefore, 1+4+9+16+25+36+49+64+81=285.
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Answer:
x=pi/3 + 2pi*k
x=2pi/3+2pi*k
Step-by-step explanation:
sin(x)=sqrt(3)/2
This happens twice in the first rotation on our unit circle.
It happens in the first quadrant and in the second quadrant. Third and fourth quadrants are negative for sine.
So we are looking for when the y-coordinate on the unit circle is sqrt(3)/2.
This is at pi/3 and 2pi/3.
So we can get all the solutions by adding +2pi*k to both of those. This gives us a full rotation about the circle any number of k times. k is an integer.
So the solutions are
x=pi/3 + 2pi*k
x=2pi/3+2pi*k
The question did not show the original vertices of A, B, and C.
Generally, for a triangle ABC that is dilated by a scale factor of four to form triangle A'B'C', the coordinates of vertices A', B', C' are:
For A(x, y)
The coordinate of vertex A' will be A' (4x, 4y)
For B(x, y)
The coordinate of vertex B' will be B' (4x, 4y)
For C(x, y)
The coordinate of vertec C' will be C' (4x, 4y)
Answer:
18
Step-by-step explanation:
100% of 6 is 6 so 300% (which does not actually exist) would be 3 times as much. 6x3=18
5.44-1.83=3.61
5._4
-_.8_ =
3.61
