ANSWER ONLY TO ASSIGNMENT QUESTION:
2 answers:
You might be interested in
Let's find the derivative of that hyperbola in order to find a slope formula to help us with this equation. We already have an x and y value. The derivative is found this way: 
so 
. The derivative supplies us with the slope formula we need to write the equation. Sub in the x value of 3 to find what the slope is: 
. So in our slope-intercept equation, x = 3, y = 1, and m = -1/3. Use these values to solve for b. 
so b = 2. The equation, then, for the line tangent to that hyperbola at that given point is 
The point L is plotted at A. (1, -1).
Step-by-step explanation:
Step 1:
First, we plot the coordinates of both triangles.
The coordinates of the triangle EFG are as follows;
E = (-4, 4), F = (-4, -2), and G = (2, -2).
The coordinates of triangle JKL are;
J = (-2, 2), K = (-2, -1), and L = (x, y).
Step 2;
Next, we need to find the distances between at least two coordinates for both triangles.
The distances between vertical lines are found by calculating the difference in y values while for horizontal lines, the distances are found by calculating the difference between the x values.
The distance of EF
The distance of JK
The distance of FG
The distance of KL = (Scale factor)(the distance of FG),
The distance of KL
Step 3;
So the point L is 3 points away from point k on the x axis.
(x, y) = (-2+3, 1) = (1, -1).
So point L is plotted at option A. (1, -1).
Answer: (1a) 250° (1b) 70° (2) see below
<u>Step-by-step explanation:</u>
1a) -110 + 360 (one rotation clockwise) = 250°
1b) 430 - 360 (one rotation counterclockwise) = 70°
2) sec Ф = -8/5 in Quadrant III
Quadrant III identifies that both sin (y) and cos (x) are negative.
sec = r/x --> r = 8, x = -5, and y = -√39
(Use Pythagorean Theorem x² + y² = r² to solve for y)
rationalized =
(GIVEN)
rationalized =
NOTE THAT YOU ARE ALLOWED A MAXIMUM OF 3 QUESTIONS PER POST. Please repost #3 and #4 as a different question and I will answer them.