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#1: "AD" is congruent to "BC" | given
<h3>Answers are:
sine, tangent, cosecant, cotangent</h3>
Explanation:
On the unit circle we have some point (x,y) such that x = cos(theta) and y = sin(theta). The sine corresponds to the y coordinate of the point on the circle. Quadrant IV is below the x axis which explains why sine is negative here, since y < 0 here.
Since sine is negative, so is cosecant as this is the reciprocal of sine
csc = 1/sin
In quadrant IV, cosine is positive as x > 0 here. So the ratio tan = sin/cos is going to be negative. We have a negative over a positive when we divide.
Because tangent is negative, so is cotangent.
The only positive functions in Q4 are cosine and secant, which is because sec = 1/cos.
Y=6.75x
He charges $6.75 per hour
The geometric term described as an infinite set of points that has
length but not width is called a line. It has a negligible with and depth. In
geometry, a line located in the plane is defined as the set of points whose
coordinates satisfy a given linear equation. A line segment however is a line
connected by two dots far apart from each other and then connected. For example
is the equation y=mx+b. This is a slope intercept form which is a linear
equation.
Answer:
The perpendicular line is:
y = 1/3 x + 2/3
Step-by-step explanation:
The given equation can be reduced to its slope-intercept form as shown below:
3x + 9y = 8y - 2
subtract 8 y from both sides
3 x + y = -2
subtract 3 x from both sides
y = - 3 x -2
therefore we know that the slope of this line is -3, and then, a perpendicular line to it must have slope given by the "opposite of the reciprocal" of this slope. That is, the slope of any perpendicular line to this one must be: 1/3
We use this slope to find the equation of the line passing through the point (13. 5)
y = 1/3 x + b
passing through (13, 5) means:
5 = 1/3 (13) + b
therefore, we can find b from the above equation
b = 5 - 13/3 = 15/3 - 13/3 = 2/3
Then the equation of this perpendicular line is:
y = 1/3 x + 2/3
