Write the equation of the line going through the points (-6,3) and (1,11). Write the final equation in slope-intercept form.
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So 1st consider that it's a square! That's very important. So for a square, all 4 sides are equal.
And now considering that the given information is the diameter. So any angle made at the circle extended from the 2 points of diameter gives an angle of 90°
Now consider one triangle. So we already know that 2 sides of the triangle are equal (because they are 2 sides of a square) , has a side of 10 (diameter) and and angle of 90°. So remaining 2 angles are 45°
Now solve it by applying
And now considering that the given information is the diameter. So any angle made at the circle extended from the 2 points of diameter gives an angle of 90°
Now consider one triangle. So we already know that 2 sides of the triangle are equal (because they are 2 sides of a square) , has a side of 10 (diameter) and and angle of 90°. So remaining 2 angles are 45°
Now solve it by applying
We have to rewrite the expression so that it has no denominator.
For example:
1 / x = x^(-1)
1/8 = 8^(-1); 1/x^(4) = x^(-4); 1/y^(3) = y^(-3); 1/z = z^(-1).
For example:
1 / x = x^(-1)
1/8 = 8^(-1); 1/x^(4) = x^(-4); 1/y^(3) = y^(-3); 1/z = z^(-1).
let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.