Please help me to find the %
1 answer:
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Answer:
Step-by-step explanation:
Consider a sketch of the problem as shown in the picture, where:
- Blue line is given by y = 4x + 1.
- Point B is the center of the circle.
- Point A is (-3, 0).
Since the center of the circle lies on the line y = 4x +1 and is tangent to the x-axis at point A, then its radius BA is perpendicular to the x-axis. To find the coordinates of point B, we must replace x = -3 into the blue line equation: y = 4x(-3) + 1 = -11.
So, we know that the center of the circle is at B=(-3, -11). And furthermore, the radius BA is of length r=11.
Since the <em>general equation of the circle</em> of radius lenght r centered at (h, k) is given by
then with h = -3, k = -11 and r= 11, the equation of our circle is
Answer:
The dimensions of the rectangular volleyball court are 60 ft x 30 ft
Step-by-step explanation:
Let
x ----> the length of rectangular volleyball court
y ---> the width of the rectangular volleyball court
we know that
The area of the rectangular volleyball court is equal to
so
----> equation A
-----> equation B
substitute equation B in equation A
Solve for y
Simplify
take square root both sides
<em>Find the value of x</em>
substitute the value of y
therefore
The dimensions of the rectangular volleyball court are 60 ft x 30 ft
9514 1404 393
Answer:
(8.49; 225°)
Step-by-step explanation:
The angle is a 3rd-quadrant angle. The reference angle will be ...
arctan(-6/-6) = 45°
In the 3rd quadrant, the angle is 45° +180° = 225°.
The magnitude of the vector to the point is its distance from the origin:
√((-6)² +(-6)²) = √(6²·2) = 6√2 ≈ 8.4859 ≈ 8.49
The polar coordinates can be written as (8.49; 225°).
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<em>Additional comment</em>
My preferred form for the polar coordinates is 8.49∠225°. Most authors use some sort of notation with parentheses. If parentheses are used, I prefer a semicolon between the coordinate values so they don't get confused with an (x, y) ordered pair that uses a comma. You need to use the coordinate format that is consistent with your curriculum materials.
