THIS IS A RIGHT & isosceles triangle: A = 90. B = 45 THEN ANGLE C =45
THEN AB = AC = 4√2
APPLY PYTHAGORAS: BC² =AB² + AC²
===> X² = (4√2)² + (4√2)² ==>16.2 +16.2 = 64
X² 64 & X =√64 ==> X = 8
Answer:
the simplified expression is: 6y-7x
Step-by-step explanation:
-y+7y= 6y
-7x
= 6y-7x
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:
The answer is A.
Step-by-step explanation:
You have to apply Pythagoras Theorem :
P.S (Take note):
Pythagoras Theorem can only applied in <u>r</u><u>i</u><u>g</u><u>h</u><u>t</u><u> </u><u>a</u><u>n</u><u>g</u><u>l</u><u>e</u><u> </u><u>t</u><u>r</u><u>i</u><u>a</u><u>n</u><u>g</u><u>l</u><u>e</u>.
Answer:
y intercept - (0, 275)
x intercept - (0, 125)
Step-by-step explanation:
