Answer:
x=2
Step-by-step explanation:
<h2>(1)</h2><h2> =(a+b)(3c-d)</h2><h2> =a(3c-d)+b(3c-d)</h2><h2> =3ac-ad+3bc-bd</h2>
<h2>(2)</h2><h2> =(a-b)(c+2d)</h2><h2> =a(c+2d)-b(c+2d)</h2><h2> =ac+2ad-bc-2bd</h2>
<h2>(3)</h2><h2> =(a-b)(c-2d)</h2><h2> =a(c-2d)-b(c-2d)</h2><h2> =ac-2ad-bc+2bd</h2>
<h2>(4)</h2><h2> =(2a+b)(c-3d)</h2><h2> =2a(c-3d)+b(c-3d)</h2><h2> =2ac-6ad+bc-3bd</h2>
<h3>Explanation:</h3>
1. PQ║TS, PQ ≅ TS, PT and QS are transversals to the parallel lines . . . given
2. ∠P ≅ ∠T . . . alternate interior angles at PT
3. ∠Q ≅ ∠S . . . alternate interior angles at QS
4. ΔPQR ≅ ΔTSR . . . ASA postulate
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You can use any pair of angles together with the sides PQ and TS. If you use the vertical angles and one of ∠T or ∠S, then you must invoke the AAS postulate for congruence, as the side is not between the two angles.
In mathematics, number sequencing of the same pattern are called progression. There are three types of progression: arithmetic, harmonic and geometric. The pattern in arithmetic is called common difference, while the pattern in geometric is called common ratio. Harmonic progression is just the reciprocal of the arithmetic sequence.
The common ratio is denoted as r. For values of r<1, the sum of the infinite series is equal to
S∞ = A₁/(1-r), where A1 is the first term of the sequence. Substituting A₁=65 and r=1/6:
S∞ = A₁/(1-r) = 65/(1-1/6)
S∞ = 78
