Wow ! There's so much extra mush here that the likelihood of being
distracted and led astray is almost unavoidable.
The circle ' O ' is roughly 98.17% (π/3.2) useless to us. The only reason
we need it at all is in order to recall that the tangent to a circle is
perpendicular to the radius drawn to the tangent point. And now
we can discard Circle - ' O ' .
Just keep the point at its center, and call it point - O .
-- The segments LP, LQ, and LO, along with the radii OP and OQ, form
two right triangles, reposing romantically hypotenuse-to-hypotenuse.
The length of segment LO ... their common hypotenuse ... is the answer
to the question.
-- Angle PLQ is 60 degrees. The common hypotenuse is its bisector.
So the acute angle of each triangle at point ' L ' is 30 degrees, and the
acute angle of each triangle at point ' O ' is 60 degrees.
-- The leg of each triangle opposite the 30-degree angle is a radius
of the discarded circle, and measures 6 .
-- In every 30-60 right triangle, the length of the side opposite the hypotenuse
is one-half the length of the hypotenuse.
-- So the length of the hypotenuse (segment LO) is <em>12 </em>.
Answer:
1=2(4)+b
Step-by-step explanation:
slope intercept looks like this y=mx(or y=#x+#
if the slope is 2 the equation will be y=2x+b
to find b so x=4 and y=1 we can plug those numbers into the equation to find b
Answer: 4 (153 - 98)
formula is a^2-b^2=(a+b)(a-b)
factor out the GCF:
4(153−98)
both terms are not perfect squares so
you cant factor it further.
x=2, y=3 is a pair of a equations right there, so a correct solution.
How about another linear system?
Answer: x+y=5, 10x + y = 23
