Pls help me .question no 14
1 answer:
Question 14, Part (i)
Focus on quadrilateral ABCD. The interior angles add to 360 (this is true for any quadrilateral), so,
A+B+C+D = 360
A+90+C+90 = 360
A+C+180 = 360
A+C = 360-180
A+C = 180
Since angles A and C add to 180, this shows they are supplementary. This is the same as saying angles 2 and 3 are supplementary.
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Question 14, Part (ii)
Let
x = measure of angle 1
y = measure of angle 2
z = measure of angle 3
Back in part (i) above, we showed that y + z = 180
Note that angles 1 and 2 are adjacent to form a straight line, so we can say
x+y = 180
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We have the two equations x+y = 180 and y+z = 180 to form this system of equations
Which is really the same as this system
The 0s help align the y terms up. Subtracting straight down leads to the equation x-z = 0 and we can solve to get x = z. Therefore showing that angle 1 and angle 3 are congruent. We could also use the substitution rule to end up with x = z as well.
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Answer:
See below for the graph.
Step-by-step explanation:
A circle or ellipse can be defined using the same sort of equation. Here, we have chosen to use the formulation ...
... ((x -a)/p)^2 +((y -b)/q)^2 -1 = 0
This will be the general form of the equation for an ellipse with center (a, b) and semi-axes p and q, in the x- and y-directions, respectively. When the axes are the same length, the ellipse is a circle.
By defining the function ...
... c(a, b, p, q, x, y) = ((x -a)/p)^2 +((y -b)/q)^2 -1
we can use the same function for all of the circles/ellipses in the figure. The parabola has vertex (0, -6) and a vertical scale factor of -1, so it can be formulated using the vertex form:
... y = k(x -a)^2 +b . . . . . for vertex (a, b) and vertical scale factor k.
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<em>The equations</em>
- (x/6)² +(y/6) -1 = 0
- (x+2)² +(y-2)² -1 = 0
- (x-2)² +(y-2)² -1 = 0
- (x/3)² +(y+2)² -1 = 0
- y = -x² -6
