Step-by-step explanation:
5b.
P (6, -1), Q (1, 3), R (x, 8)
the distance between points is found via Pythagoras for right-angled triangles, as the distance is the Hypotenuse c, and the coordinate differences of x and y are the "legs".
for PQ
c² = (6-1)² + (-1 - 3)² = 5² + (-4)² = 25 + 16 = 41
now we need find the x, so that we get 41 as the square of the distance for QR
41 = (1-x)² + (3-8)² = (1-x)² + (-5)² = (1-x)² + 25
aha ! the 41 and 25 terms are the same as above.
that means (1-x)² must be equal to 16.
=>
1-x = 4 => x = -3
or
1-x = -4 => x = 5
so, R could be either (-3, 8) or (5, 8).
5c.
(4 + sqrt(7))/(2×sqrt(7) + 5) = a + b×sqrt(7)
let's use the old trick of
(a+b)(a-b) = a² - b²
to get rid of the sqrt(7) in the denominator.
we multiply with
(2×sqrt(7) - 5)/(2×sqrt(7) - 5)
=>
(4 + sqrt(7))×(2×sqrt(7) - 5) / (2×sqrt(7) + 5)×(2×sqrt(7) - 5) =
= (8×sqrt(7) - 20 +2×7 - 5×sqrt(7)) / (4×7 - 25) =
= (3×sqrt(7) - 6) / 3 = sqrt(7) - 2
now, we see
sqrt(7) - 2 = a + b×sqrt(7)
a = the sum of every term without sqrt(7) = -2
b = the sum of all factors of sqrt(7) = 1
6a.
similar triangles.
EBA and DFA are similar triangles due to the parallel baselines and the sharing of the same "legs" and therefore using the same angles.
and that means that all the lengths of lines in one triangle are the result of the lengths of the corresponding lines in the other triangle multiplied by the same factor.
since E is the midpoint of AD, that means this factor is 2 :
AE×2 = AD.
therefore, FA = AB×2 and AB = FA/2
that proves that B is the midpoint of FA.
that also means that L is the midpoint of DF.
and DF = EB×2 or EB = DF/2
as L is the midpoint of DF, it means that LF = DF/2 = EB.
6b.
x - 2/x = 5
i.
let's square everything.
(x - 2/x)² = 25
x² - 2×2x/x + 4/x² = 25
x² - 4 + 4/x² = 25
x² + 4/x² = 29
ii.
let's multiply with (x - 2/x) again.
(x² + 4/x²)×(x - 2/x) = 29×5
x³ - 2x²/x + 4x/x² - 8/x³ = 145
x³ - 2x + 4/x - 8/x³ = 145
x³ - 8/x³ - 2×(x - 2/x) = 145
x³ - 8/x³ - 2×5 = 145
x³ - 8/x³ = 155
6c.
I can't draw here, but let's transform the equations into regular line equations :
2y - x = 8
2y = x + 8
y = (1/2)x + 4
when assuming x = 0 for one point. and y = 0 for a second point we get (0, 4) and (-8, 0)
y - 2x = 1
y = 2x + 1
same trick as above with x = 0 and y = 0 :
(0, 1) and (-1/2, 0). but we could also use x=1 to get an integer as x coordinate for the second point: (1, 3)
7a.
i.
the radius is half the diameter = (18+8)/2 = 13 cm
ii.
the shortest chord through M creates a right-angled triangle with the radius from 0 to the point on the circle, where the chord ends as the Hypotenuse. one leg is M0, which is
13 - 8 = 5 cm
the other leg is half of the length of the chord, as the chord goes up and down of AB.
so,
13² = 5² + x²
169 = 25 + x²
144 = x²
x = 12 cm
so, the whole chord is 2×12 = 24cm long.
7b.
3/(sqrt(6) + sqrt(3)) rationalized is
3×(sqrt(6) - sqrt(3)) / (sqrt(6) + sqrt(3))×(sqrt(6) - sqrt(3))
3×(sqrt(6) - sqrt(3)) / (6 - 3) =
3×(sqrt(6) - sqrt(3)) / 3 = sqrt(6) - sqrt(3)
1/(sqrt(3) + sqrt(2)) rationalized is
(sqrt(3) - sqrt(2)) / (sqrt(3) + sqrt(2))×(sqrt(3) - sqrt(2))
(sqrt(3) - sqrt(2)) / (3 - 2) = sqrt(3) - sqrt(2)
4/(sqrt(6) + sqrt(2)) rationalized is
4×(sqrt(6) - sqrt(2)) / (sqrt(6) + sqrt(2))×(sqrt(6) - sqrt(2))
4×(sqrt(6) - sqrt(2)) / (6 - 2) =
4×(sqrt(6) - sqrt(2)) / 4 = sqrt(6) - sqrt(2)
so, we have in total
sqrt(6) - sqrt(3) + sqrt(3) - sqrt(2) - sqrt(6) + sqrt(2) = 0
the whole expression is eliminating every term and is simply 0.
