Solve for the width in the formula for the area of a rectangle.
2 answers:
You might be interested in
Part A. The technique on how to find the equation that only applies to point D and E, is to create a line or curve that only includes two of these points. In this case, I created a random parabola that isolates points C and F from the rest of the points. First, we have to find the equation of the parabola through its general forms:
(x - h)² = +/-4a(y-k) or (y - k)² = +/-4a(x - h)
For parabolas drawn like that in the picture, the general form is (x - h)² = +4a(y-k), where the vertex is (h,k) and a is the distance from the vertex to the focus. From the picture, the vertex is at (0,3). Then, we use point D(-2,4) to determine a:
(-2 - 0)² = +4a(4 - 3)
4a = 4
So, the equation of the parabola is:
x² = 4(y - 3)
x² = 4y - 12
Part B. Point D was already verified above. Now for point E(2,4)
x² = 4y - 12
2² ? 4(4) - 12
4 ? 4
4 = 4
Part C. For y < 7x − 4, ignore the equality symbol first and graph the line. Assign random values of x, then you get corresponding values of y. Plot them as shown in the second picture. The line is shown in red. Next, test the equation by choosing a random point. Let's choose the purple point at (4,3).
3 ? 7(4) − 4
3 ? 24
3 < 24
Thus, it applies to the purple point, and all the other areas to that area. The shaded region are all solutions of the inequality. So, Erica is only interested in points E, C and F.
(x - h)² = +/-4a(y-k) or (y - k)² = +/-4a(x - h)
For parabolas drawn like that in the picture, the general form is (x - h)² = +4a(y-k), where the vertex is (h,k) and a is the distance from the vertex to the focus. From the picture, the vertex is at (0,3). Then, we use point D(-2,4) to determine a:
(-2 - 0)² = +4a(4 - 3)
4a = 4
So, the equation of the parabola is:
x² = 4(y - 3)
x² = 4y - 12
Part B. Point D was already verified above. Now for point E(2,4)
x² = 4y - 12
2² ? 4(4) - 12
4 ? 4
4 = 4
Part C. For y < 7x − 4, ignore the equality symbol first and graph the line. Assign random values of x, then you get corresponding values of y. Plot them as shown in the second picture. The line is shown in red. Next, test the equation by choosing a random point. Let's choose the purple point at (4,3).
3 ? 7(4) − 4
3 ? 24
3 < 24
Thus, it applies to the purple point, and all the other areas to that area. The shaded region are all solutions of the inequality. So, Erica is only interested in points E, C and F.
Answer:
Part 4)
Part 10) The angle of elevation is
Part 11) The angle of depression is
Part 12) or
Part 13) or
Step-by-step explanation:
Part 4) we have that
The angle theta lies in Quadrant II
so
The sine of angle theta is positive
Remember that
substitute the given value
Part 10)
Let
----> angle of elevation
we know that
----> opposite side angle theta divided by adjacent side angle theta
Part 11)
Let
----> angle of depression
we know that
----> opposite side angle theta divided by hypotenuse
Part 12) What is the exact value of arcsin(0.5)?
Remember that
therefore
-----> has two solutions
----> I Quadrant
or
----> II Quadrant
Part 13) What is the exact value of
The sine is negative
so
The angle lies in Quadrant III or Quadrant IV
Remember that
therefore
----> has two solutions
----> IV Quadrant
or
----> III Quadrant
We know the cofunction identity 
.
Thus, the tangent of 90° minus <span>θ also equals 5.</span>
Thus, the tangent of 90° minus <span>θ also equals 5.</span>