Answer:
Part A: Angle R is not a right angle.
Part B; Angle GRT' is a right angle.
Step-by-step explanation:
Part A:
From the given figure it is noticed that the vertices of the triangle are G(-6,5), R(-3,1) and T(2,6).
Slope formula
The product of slopes of two perpendicular lines is -1.
Slope of GR is
Slope of RT is
Product of slopes of GR and RT is
Therefore lines GR and RT are not perpendicular to each other and angle R is not a right angle.
Part B:
If vertex T translated by rule
Then the coordinates of T' are
Slope of RT' is
Product of slopes of GR and RT' is
Since the product of slopes is -1, therefore the lines GR and RT' are perpendicular to each other and angle GRT' is a right angle.
F(x)=(-2/((x+y-2)^(1/2))-(x+y+2)^(1/2)
the only irrational part of this expression is the (x+y-2)^(1/2) in the denominator, so, to rationalize this, you multiply the numerator and denominator by the denominator, as well as the other parts of the expression
also, you must multiply the -sqrt(x+y+2) by sqrt(x+y-2)/sqrt(x+y-2) to form a common denominator
(-2)/(x+y-2)^(1/2)-(x+y+2)^(1/2)(x+y-2)^(1/2)/(x+y-2)^(1/2)
(common denominator)
(-2-(x^2+xy+2x+xy+y^2+2y-2x-2y-4))/(x+y-2)^(1/2)
(FOIL)
(-2-x^2-y^2-2xy+4)/(x+y-2)^(1/2)
(Distribute negative)
(-x^2-y^2-2xy+2)/(x+y-2)^(1/2)
(Simplify numerator)
(-x^2-y^2-2xy+2)(x+y-2)^(1/2)/(x+y-2)^(1/2)(x+y-2)^(1/2)
(Rationalize denominator by multiplying both top and bottom by sqrt)
(-x^2-y^2-2xy+2)((x+y-2)^(1/2))/(x+y-2)
(The function is now rational)
=(-x^2-y^2-2xy+2)(sqrt(x+y-2))/(x+y-2)
The answer would be 1/6 assuming a fair trial
Answer is c la diferencia Ed la Miya’s question el minuendo
Answer:
Probably 10
Step-by-step explanation:
10/5=2
