Help please tap the picture above

Mathematics

1 answer:

Wewaii [24]2 years ago

7 0

Answer:

a = 8

r = 15%

When t = 3, y= 4.9

Step-by-step explanation:

8 represents the original amount because that's what you're increasing, so a=8.

To find your percentage, look into the parenthesis to see the decimal. We see that it's 0.15, but that's not a percentage yet, so you have to multiply it by 100. Or, moving your decimal two times to the right. Therefore, r=15%.

Now to find what the function will equal when t=3, simply plug 3 for t. I've always used a calculator for this so I'm not going to write it out (sorry). Your result should be 4.9 (this is already rounded to the nearest tenth for you).

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Alexus [3.1K]

Answer:

Step-by-step explanation:

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6 0

3 years ago

Joey has $400 in a bank account that earns 8% in annual interest. Would this equation be Linear or Exponential?

Dennis_Churaev [7]

Exponential

The first year, he would earn 8% interest on the $400 deposited, so he'd have $432. The next year, he would earn 8% interest on the interest earned the previous year in addition to the amount deposited, so he'd have $466.56.

His bank balance would follow the compound interest equation which is exponential.

= 400 × 1.08^t

where t is the number of years since the deposit

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3 years ago

[2/SEC(¶/3)•[lim x→0 x^3+8x+10]^2]/[lim θ→0 sinθ/θ]

Oliga [24]

Assuming the pilcrow is supposed to represent pi:

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100

4 0

3 years ago

What is the sum of the geometric series?

Dvinal [7]

Essentially this is asking you to sum the first three values of the geometric series that starts with i being 1 and has the rule of ( $8(\frac{1}{4})^{i-1}$ ).

Since we must sum the first <em>three </em>geometric values we need to see what the value of i will be for the first three. The sigma notation shows us that the first i will be equal to 1. This means the second i is 2 and the third i is 3.

Knowing this you can plug in the corresponding i values into ( $8(\frac{1}{4})^{i-1}$ ) and sum it all together

( $8(\frac{1}{4})^{1-1}$ ) + ( $8(\frac{1}{4})^{2-1}$ ) + ( $8(\frac{1}{4})^{3-1}$ )