A jar contains 30 red marbles, 50 blue marbles, and 20 white marbles. You choose one marble from the jar at random. What is the
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Answer:
Step-by-step explanation:
Lets add +5 in both the sides.
Now lets divide 12 on both the sides.
For this fraction, you would have to see how many times 7 goes into 47. So, 6 times as 7 times 6 is 42. Subtract 42 from 47, and this would be your remainder:
6
6
The general equation for a circle, 
, falls out of the Pythagorean Theorem, which states that the square of the hypotenuse of a right triangle is always equal to the sum of the squares of its legs (you might have seen this fact written like 
, where <em>a </em>and <em>b</em> are the legs of a right triangle and <em>c </em>is its hypotenuse. When we fix <em /><em>c</em> in place and let <em>a </em>and <em>b </em>vary (in a sense, at least; their values are still dependent on <em>c</em>), the shape swept out by all of those possible triangles is a circle - a shape defined by having all of its points equidistant from some center.
How do we modify this equation to shift the circle and change its radius, then? Well, if we want to change the radius, we simply have to change the hypotenuse of the triangle that's sweeping out the circle in the first place. The default for a circle is 1, but we're looking for a radius of 6, so our equation, in line with Pythagorus's, would look like
, or 
.
Shifting the center of the circle is a bit of a longer story, but - at first counterintuitively - you can move a circle's center to the point (a,b) by altering the x and y portions of the equation to read:
How do we modify this equation to shift the circle and change its radius, then? Well, if we want to change the radius, we simply have to change the hypotenuse of the triangle that's sweeping out the circle in the first place. The default for a circle is 1, but we're looking for a radius of 6, so our equation, in line with Pythagorus's, would look like
Shifting the center of the circle is a bit of a longer story, but - at first counterintuitively - you can move a circle's center to the point (a,b) by altering the x and y portions of the equation to read: