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 ispoint (-5,12)
see the attached figure
Answer:
Step-by-step explanation:
tell me the question ?? did not say the question ok ??
Answer with explanation:
A regular polygon is simple closed geometrical shape, only made up of line segments.
And Regular Tessellation, is geometrical shape, made up of regular polygons having, 3 ,or 4 or 6 sides.
It is combination of, only single kind of polygon,either, polygon having 3 sides, or ,4 sides or 6 sides.
→Regular polygon having 3 sides is Called Equilateral Triangle.
→Regular polygon having 4 sides is Called Square.
→Regular polygon having 6 sides is Called Regular Hexagon.
⇒Regular polygon having , 8 sides will not form a regular tessellation.
Option D
Answer:
<h3>perpendicular line:
y = -¹/₆
x + 4¹/₃
</h3><h3> parallel line:
y = 6x - 45
</h3>
Step-by-step explanation:
y=m₁x+b₁ ⊥ y=m₂x+b₂ ⇔ m₁×m₂ = -1
{Two lines are perpendicular if the product of theirs slopes is equal -1}
y = 6x - 7 ⇒ m₁=6
6×m₂ = -1 ⇒ m₂ = -¹/₆
The line y=-¹/₆
x+b passes through point (8, 3) so the equation:
3 = -¹/₆
×8 + b must be true
3 = -⁴/₃ + b
b = 4¹/₃
Therefore equation in slope-intercept form:
y = -¹/₆
x + 4¹/₃
y=m₁x+b₁ ║ y=m₂x+b₂ ⇔ m₁ = m₂
{Two lines are parallel if their slopes are equal}
y = 6x - 7 ⇒ m₁=6 ⇒ m₂=6
The line y=6x+b passes through point (8, 3) so the equation:
3 = 6×8 + b must be true
3 = 48 + b
b = -45
Therefore equation in slope-intercept form:
y = 6x - 45
If the drawing of your octagon (or whatever) has been separated into triangles, and one triangle's area<span> is labeled, then you do not need to know the apothem. Just take the </span>area<span> of that one triangle, and multiply by the number of sides in the original </span>polygon<span>.</span>
