Consider the figure shown below. Find the value of each variable mentioned in the figure below​

Consider the figure shown below. Find the value of each variable mentioned
3 weeks ago

Solution 1

Guest Guest #7253155
3 weeks ago
  • a = 26, b = 64, c = 42

-------------------

The measure of an inscribed angle is half the measure of the intercepted arc:

  • a = 52/2 = 26,
  • c = 84/2 = 42.

Angles a and b sum to a right angle since inscribed between the tangent and diameter.

Then we have:

  • a + b = 90
  • 26 + b = 90
  • b = 90 - 26
  • b = 64

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Fred deposits $6,500 in a saving account that pays 0.5% interest, compounded quarterly. Round each answer to the nearest cent.
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b. Find the first quarter's ending balance.
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The amount of interest earned quarterly at 0.5% interest rate compounded quarterly are;

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The amount Fred deposited, P = $6,500

The interest paid by the account, r = 0.5% compounded quarterly

a. The first quarter interest can be found by the formula;

[tex]C.I. = P\cdot \left(1 + \dfrac{r}{4} \right)^{4\times \dfrac{1}{4} }- P[/tex]

Therefore;

[tex]C.I. = 6500\times \left(1 + \dfrac{\frac{0.5}{100} }{4} \right)^{4\times \dfrac{1}{4} }- 6500 = 8.125[/tex]

The interest after the first period is $8.125

b. The first quarter ending balance is therefore;

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A = $6,500 + $8.125 = $6,508.125

c. The second quarter's interest can be found from the formula;

[tex]C.I. = P\cdot \left(1 + \dfrac{r}{4} \right)^{4\times t} - P[/tex]

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[tex]C.I. = 6508.125\times \left(1 + \dfrac{\frac{0.5}{100} }{4} \right)^{4\times \dfrac{1}{4} }- 6508.125 = 8.13515625[/tex]

The second quarter's interest is $8.13515625

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A = $6,508.125 + $8.13515625 = $6,516.26015625

The second quarter ending balance is $6,516.26015625

e. The third quarters interest is therefore;

[tex]C.I. = 6516.26015625\times \left(1 + \dfrac{\frac{0.5}{100} }{4} \right)^{4\times \dfrac{1}{4} }- 6516.26015625 = 8.1432520325[/tex]

The third quarter's interest is about $8.1433

f. The third quarter's ending balance is therefore;

A ≈ $6516.26015625 + $8.1432520325 = $6,524.40548145

The third quarter's ending balance is $6,524.40548145

g. The fourth quarter's interest can be found as follows;

[tex]C.I. = 6,524.40548145 \times \left(1 + \dfrac{\frac{0.5}{100} }{4} \right)^{4\times \dfrac{1}{4} }- 6524.40548145 = 8.15550685975[/tex]

The fourth quarter's interest is; $8.15550685975

h. The balance at the end of one year is therefore;

Balance = C.I. fourth quarter + Principal for the fourth quarter

Therefore;

Balance = $6,524.40548145 + $8.15550685975 = $6,532.56098831

i. The interest the account earns in a year is therefore;

$6,532.56098831 - $6,500 = $32.5609883

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