There is a line and a parabola in the graph.
So, we will get the solution set from the point of intersection of both line and parabola.
Notice that the parabola and the line intersecting at two points E(1, 5) and C(-0.5,2).
So, the solution set is E(1, 5) and C(-0.5,2).
The solutions appear to be
{π/2, 2π/3, 4π/3}.
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Replacing sin(2x) with 2sin(x)cos(x), you have
2sin(x)cos(x) +sin(x) -2cos(x) -1 = 0
sin(x)(2cos(x) +1) -(2cos(x) +1) = 0 . . . . factor by grouping
(sin(x) -1)(2cos(x) +1) = 0
This has solutions
sin(x) = 1
x = π/2and
2cos(x) = -1
cos(x) = -1/2
x = {2π/3, 4π/3}
Explanation:
All of these are done the same way.
1. Set your compass to a distance slightly longer than the distance from the point to the line you want the perpendicular to.
2. Using the point as a center, draw an arc that intersects the line in 2 places. Consider those points to be "P" and "Q".
3. You can leave the compass as is, or set it to any other distance longer than half the distance between P and Q. Using this radius, and using P and Q as centers, draw intersecting arcs on the other side of the line from the original point. Consider the intersection of those arcs to be point "R".
4. A line through the original point and point R will be perpendicular to the line, as you want.
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In problem 1, the point and line are obvious.
In problem 2, the point is the vertex opposite the line of interest. (There will be 3 similar constructions.)
Hello,
1) we imagine a linear function:
y=ax+b
(1,3)==>3=a+b (1)
(2,6)==>6=2a+b (2)
(2)-(1)==>a=3==>b=3-3=0
So we find y=3x
But (3,11)==>11=3*3 is false ==> no linear function A,B and D are false.
2) we imagine a quadratic function
y=ax²+bx+c
(1,3)==>3=a+b+c (1)
(2,6)==>6=4a+2b+c (2)
(3,11)==>11=9a+3b+c (3)
(3)-(2)==>5a+b=5 (4)
(2)-(1)==>3a+b=3 (5)
(4)-(5)==> 2a=2==> a=1
==> b=5-5*1=0
(1)==>3=1+0+c==>c=2
The quadratic function is y=x²+2
and you have make a mistake in your question!
Answer is C if you change in y=x²+2
Proof:
(4,18) 18=1*4²+0*4+2=16+2=18
