Draw a rectangle with diagonal 5 in. Inside this rect. are 2 acute triangles of hypotenuse 5. Note that 3^2 + 4^2 = 5^2; thus the width of the rect. is 3 and the length is 4, with the result that the hypo. is sqrt(3^2+4^2), as expected.
9514 1404 393
Answer:
5) 729, an=3^n, a[1]=3; a[n]=3·a[n-1]
6) 1792, an=7(4^(n-1)), a[1]=7; a[n]=4·a[n-1]
Step-by-step explanation:
The next term of a geometric sequence is the last term multiplied by the common ratio. (This is the basis of the recursive formula.)
The Explicit Rule is ...

for first term a₁ and common ratio r.
The Recursive Rule is ...
a[1] = a₁
a[n] = r·a[n-1]
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5. First term is a₁ = 3; common ratio is r = 9/3 = 3.
Next term: 243×3 = 729
Explicit rule: an = 3·3^(n-1) = 3^n
Recursive rule: a[1] = 3; a[n] = 3·a[n-1]
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6. First term is a₁ = 7; common ratio is r = 28/7 = 4.
Next term: 448×4 = 1792
Explicit rule: an = 7·4^(n-1)
Recursive rule: a[1] = 7; a[n] = 4·a[n-1]
The sum of angles of a triangle is 180°, so m∠K = 180° -45° -30° = 105°.
The Law of Sines tells you
... FL/sin(∠K) = FK/sin(∠L)
Solving for FL, we get
... FL = FK·sin(∠K)/sin(∠L)
... FL = a·sin(105°)/sin(30°) = a·sin(105°)/(1/2)
... FL = 2a·sin(105°) ≈ 1.93185a
312 I think, hop[e this helps
It would b 9 x (-3^(x-1))