(High points) please solve with explanation. Legit answers only thank you!
1 answer:
Step-by-step explanation:
The x-intercept is at (-3, 0). To find the slope of the perpendicular line, let's rewrite the equation to its slope-intercept form, which is
y = (2/3)x + 2
This means that the perpendicular line has a slope m = -3/2. So the slope-intercept form of this new line is
y = -(3/2)x + k. Since this line passes through (-3, 0), we cal solve for k:
0 = -(3/2)(-3) + k or k = -9/2
So our equation becomes (slope-intercept form)
or in its standard form,
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We have that
csc ∅=13/12
sec ∅=-13/5
cot ∅=-5/12
we know that
csc ∅=1/sin ∅
sin ∅=1/ csc ∅------> sin ∅=12/13
sec ∅=-13/5
sec ∅=1/cos ∅
cos ∅=1/sec ∅------> cos ∅=-5/13
sin ∅ is positive and cos ∅ is negative
so
∅ belong to the II quadrant
therefore
<span>the coordinates of point (x,y) on the terminal ray of angle theta are
</span>x=-5
y=12
the answer is
point (-5,12)
see the attached figure
csc ∅=13/12
sec ∅=-13/5
cot ∅=-5/12
we know that
csc ∅=1/sin ∅
sin ∅=1/ csc ∅------> sin ∅=12/13
sec ∅=-13/5
sec ∅=1/cos ∅
cos ∅=1/sec ∅------> cos ∅=-5/13
sin ∅ is positive and cos ∅ is negative
so
∅ belong to the II quadrant
therefore
<span>the coordinates of point (x,y) on the terminal ray of angle theta are
</span>x=-5
y=12
the answer is
point (-5,12)
see the attached figure