Answer:
The answer is below
Step-by-step explanation:
A parallelogram is a quadrilateral (has 4 sides and 4 angle) with two pair of parallel and opposite sides. Opposite sides of a parallelogram are parallel and equal.
Given parallelogram ABCD:
AB = CD = 18 cm; BC = AD = 8 cm
∠P = ∠P, ∠PDA = ∠PCQ (corresponding angles are equal).
Hence ΔPCQ and ΔPDA are similar by angle-angle similarity theorem. For similar triangles, the ratio of their corresponding sides equal. Therefore:
Perimeter of CPQ = CP + CQ + PQ
15 = 6 + 8/3 + PQ
PQ = 15 - (6 + 8/3)
PQ = 6.33
∠CQP = ∠AQB (vertical angles), ∠QCP = ∠QBA (alternate angles are equal).
Hence ΔCPQ and ΔABQ are similar by angle-angle similarity theorem
Perimeter of BAQ = AB + BQ + AQ = 18 + 8 + 19 = 45cm
PA = AQ + PQ = 19 + 6.33 = 25.33
PD = CD + DP = 18 + 6 = 24
Perimeter of PDA = PA + PD + AD = 24 + 25.33 + 8 = 57.33 cm
The linear equation:
y = k x + b
We will use first two values and solve the system of equations:
11 = k * 7 + b / *(-1)
13 = k * 8 + b
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-11 = - k * 7 - b
13 = k * 8 + b
----------------------
2 = k
13 = 16 + b
b = 13 - 16 = -3
y = 2 x - 3 ( y varies directly with x )
Answer:
Numerator = 2(b^2+a^2) or equivalently 2b^2+2a^2
Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3
Step-by-step explanation:
Let
S = 2b/(b+a)^2 + 2a/(b^2-a^2) factor denominator
= 2b/(b+a)^2 + 2a/((b+a)(b-a)) factor denominators
= 1/(b+a) ( 2b/(b+a) + 2a/(b-a)) find common denominator
= 1/(b+a) ((2b*(b-a) + 2a*(b+a))/((b+a)(b-a)) expand
= 1/(b+a)(2b^2-2ab+2ab+2a^2)/((b+a)(b-a)) simplify & factor
= 2/(b+a)(b^2+a^2)/((b+a)(b-a)) simplify & rearrange
= 2(b^2+a^2)/((b+a)^2(b-a))
Numerator = 2(b^2+a^2) or equivalently 2b^2+2a^2
Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3
5/6x -4 = -2
5/6x = 2
multiply both sides by 6 to get rid of the fraction.
5x = 12
x = 2.4
Assuming it’s out of 100 it would 70 games they have won, but if it provides a number you will need to provide that for the correct answer.
