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Ad libitum [116K]
3 years ago
14

Find the value of an investment of $15,000 for 13 years at an annual interest rate of 4.55% compounded continuously.

Mathematics
1 answer:
atroni [7]3 years ago
4 0

Answer:

A = $27100.5

Step-by-step explanation:

We are given;

Principal: P = $15,000

Interest rate: r = 4.55% = 0.0455

Time in years: t = 13 years

The formula for the value of the investment after t years of continuous compounding is;

A = P(e^(rt))

A = 15000(e^(0.0455 × 13))

A = 15000 × 1.8067

A = $27100.5

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Put the following equation in standard form and determine the quadratic, linear, and constant coefficients. -3x2 - 8 = 5x - 7
marusya05 [52]

Answer:

  • -3x² -5x -1 = 0
  • -3 (quadratic)
  • -5 (linear)
  • -1 (constant)

Step-by-step explanation:

The equation will be in standard form when terms are listed in order of decreasing degree, and the right side of the equation is 0.

<h3>Standard form</h3>

We can subtract the right-side expression from both sides to get standard form.

  -3x² -8 -(5x -7) = 5x -7 -(5x -7)

  -3x² -8 -5x +7 = 0 . . . . . . . simplify a bit

  -3x² -5x -1 = 0 . . . . . . . . . collect terms

The standard form equation can be written ...

  -3x² -5x -1 = 0

<h3>Coefficients</h3>

The quadratic coefficient is the coefficient of the term with degree 2. The quadratic coefficient is -3.

The linear coefficient is the coefficient of the term with degree 1. The linear coefficient is -5.

The constant coefficient is the coefficient of the term with no variables. The constant is -1.

__

<em>Additional comment</em>

We can make the leading coefficient positive by multiplying the equation by -1. This gives ...

  3x² +5x +1 = 0

with quadratic, linear, and constant coefficients 3, 5, 1.

This is a legitimate answer to this question. In the case of linear equations, the "standard form" has the constant on the right side of the equal sign, and the leading coefficient is required to be positive. A negative leading coefficient can sometimes lead to errors (when the sign is overlooked), so having a positive leading coefficient is often preferred.

8 0
2 years ago
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