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blagie [28]
2 years ago
9

Given the following exponential function, identify whether the change represents growth or decay, and determine the percentage r

ate of increase or decrease. y=38(1.108)^x
Mathematics
2 answers:
Vitek1552 [10]2 years ago
3 0

Step-by-step explanation:

We are given this function:

y=8700* (1.04)^{4}y=8700∗(1.04)

4

8700 is the initial amount.

1.04 shows the change of original amount. This is decimal form of percentage. We need to transform it into regular percentage.

1.04 * 100% = 104%

Now we observe this number. If it is greater than 100% we have growth, if it is lower than 100% it is decay, and if it is equal to 100% than there is no change.

In our case this number is greater than 100% so we have growth. To determine the percentage rate we must substract 100% as it represents the original amount.

104% - 100% = 4%

This would be our solution if we don't have an exponent.

We have exponent so first step is to calculate the number and then we repeat the steps from above.

1.04^{4} = 1,169858561.04

4

=1,16985856

1,16985856 * 100% ≈ 116,99%

116.99% - 100% = 16.99%

So, final solution is growth of 16.99%

Fudgin [204]2 years ago
3 0

We are given this function:

y=8700* (1.04)^{4}y=8700∗(1.04)

4

8700 is the initial amount.

1.04 shows the change of original amount. This is decimal form of percentage. We need to transform it into regular percentage.

1.04 * 100% = 104%

Now we observe this number. If it is greater than 100% we have growth, if it is lower than 100% it is decay, and if it is equal to 100% than there is no change.

In our case this number is greater than 100% so we have growth. To determine the percentage rate we must substract 100% as it represents the original amount.

104% - 100% = 4

This would be our solution if we don't have an exponent.

We have exponent so first step is to calculate the number and then we repeat the steps from above.

1.04^{4} = 1,169858561.04

4

=1,16985856

1,16985856 * 100% ≈ 116,99%

116.99% - 100% = 16.99%

So, final solution is growth of 16.99%

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