N + d = 21
0.5n + 0.10d = 1.35
5n + 10d = 135
d = 21-n
5n + 210 - 10n = 135
-5n = -75
n = 15
d = 6
The recursive sequence that would produce the sequence 8,-35,137,… is T(n + 1) = -3 - 4T(n) where T(1) = 8
<h3>How to determine the recursive sequence that would produce the sequence?</h3>
The sequence is given as:
8,-35,137,…
From the above sequence, we can see that:
The next term is the product of the current term and -4 added to -3
i.e.
Next term = -3 + Current term * -4
So, we have:
T(n + 1) = -3 + T(n) * -4
Rewrite as:
T(n + 1) = -3 - 4T(n)
Hence, the recursive sequence that would produce the sequence 8,-35,137,… is T(n + 1) = -3 - 4T(n) where T(1) = 8
Read more about recursive sequence at
brainly.com/question/1275192
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to write 98 as a product of its prime factors we have to first find the prime factors of 98
prime factors are prime numbers by which the given number can be divided by.
98 we have to keep dividing it by prime numbers
98 is an even number so we can first divide by 2
98 / 2 = 49
49 is a multiple of 7 which too is a prime number so we can divide 49 by 7
49/7 = 7
7 can be divided again by 1
7/7 = 1
98 is divisible by 2 and 7
so 98 written as a product of prime factors is
98 = 2 x 7 x 7
Answer:
see explanation
Step-by-step explanation:
(a)
(the side required )² = sum of squares of other 2 sides - ( 2 × product of other 2 sides and cos(angle opposite side required ) )
x² = 12² + 15² - (2 × 12 × 15 × cos71°)
(b)
x² = 144 + 225 - 360cos71°
= 369 - 360cos71° ( take square root of both sides )
x = 
≈ 16 cm ( to the nearest integer )
