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frez [133]
3 years ago
13

How do I make an algebraic formula for a number that's increasing by 74%?

Mathematics
1 answer:
Ronch [10]3 years ago
7 0

Answer:

Step-by-step explanation:

If the number is increasing at a constant rate of 74%, this rate of increment is in geometric progression. The formula for the nth term of a geometric sequence can be expressed as

Tn = ar^(n - 1)

Where

n represents the number of terms in there sequence.

a represents the first term of the sequence.

r represents the common ratio

From the information given,

The number increases by 74%, it means that

r = 1 + 74/100 = 1 + 0.74 = 1.74

Starting number is 93.13 so

a = 93.13

It would increase 100 times so

n = 100

Therefore, the formula will be

Tn = 93.13 × 1.74^(n - 1)

When n = 100,

T100 = 93.13 × 1.74^(100 - 1)

T100 = 93.13 × 1.74^99

T100 = 6521.92 × 10^20

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Answer:

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<u>Or</u><u> </u><u>;</u> RT=RS+ST =75+28= 103

<h3><u>So</u><u>;</u></h3><h3>a. <u>Ans;</u> y=8</h3>

<h3>b. <u>Ans;</u> RS= 75 , ST = 28 , RT = 103</h3>

I hope I helped you^_^

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