If the measure of arc DEF is 248°, what is the measure of ∡DEF?
2 answers:
the complete question in the attached figure
we know that
angle < DEF is an inscribed angle.
An inscribed angle is an angle with its vertex on the circle which is formed by two intersecting chords.
The inscribed angle measures half of the arc it comprises
the arc it comprises is the arc DCF
the arc DCF is equal to
Hence
∡DEF is equal to arc DCF/2
the answer is
∡DEF=
°
we know that
angle < DEF is an inscribed angle.
An inscribed angle is an angle with its vertex on the circle which is formed by two intersecting chords.
The inscribed angle measures half of the arc it comprises
the arc it comprises is the arc DCF
the arc DCF is equal to
Hence
∡DEF is equal to arc DCF/2
the answer is
∡DEF=
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The solution to these equations is (x, y) = (2, 4).
We've not heard of a method of solving a linear system using the x- and y-intercepts. A Google search on the subject turns up this question, and no other information. However, there is information you can gain from the intercepts that can help you find a solution. (Your reference material may provide a better source of information on this subject.)
The <em>intercept form</em> of a linear equation is ...
... x/(x-intercept) + y/(y-intercept) = 1
Dividing the first equation by 4, you can rearrange it to ...
... x/(-2) +y/(2) = 1 . . . . . . the x-intercept is -2, the y-intercept is +2.
Dividing the second equation by -6, you can rearrange it to ...
... x/-6 +y/3 = 1 . . . . . . . . the x-intercept is -6, the y-intercept is +3.
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<em>What you can do with the intercepts</em>
The intercepts can be used to <em>graph the equations</em>. Plot each of the intercepts for a given equation, then draw a line through them. (See the attachment.)
Here, both lines have their intercepts on the negative x and positive y axes. The slopes and intercepts of the lines are such that they intersect in the 1st quadrant.
We can use the intercepts to <em>find the slope of the line</em>.
The slope of each line can be found from ...
... slope = -(y-intercept)/(x-intercept)
Then the slope of the first line is ...
... m1 = -2/-2 = 1
and the slope of the second line is ...
... m2 = -3/-6 = 1/2
The difference in slopes is ...
... m1 - m2 = 1 - 1/2 = 1/2
Using the slope and intercept we can <em>find the solution</em> by substitution or elimination.
For line 1 with slope m1 and y-intercept b1, the equation of the line in slope-intercept form is
... y = m1·x + b1 = x +2
For line 2 with slope m2 and y-intercept b2, the equation of the line in slope-intercept form is
... y = m2·x +b2 = (1/2)x +3
Subtracting the second equation from the first eliminates the y-variable and gives ...
... y - y = 0 = (x +2) -(1/2x +3) = 1/2x - 1
Adding 1 and multiplying by 2 gives the solution for x:
... x = 2
Then the first equation gives the solution for y:
... y = x + 2 = 2+2 = 4
The solution is (x, y) = (2, 4).
_____
<em>Comment on </em><em>find the solution</em>
Above, we subtracted one equation from the other to get ...
... y - y = 0 = x(m1 -m2) +(b1 -b2)
We could have simply equated the values of y to get ...
... m1·x +b1 = y = m2·x +b2
Either way you do this, you find the solution for x is ...
... x = (b2 -b1)/(m1 -m2)
