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kramer
2 years ago
14

If $4,000 is invested in an account for 25 years. Calculate the total interest earned at the end of 25 years if the interest is:

(a) 7% simple interest: $ (b) 7% compounded annually: $ (c) 7% compounded quarterly: $ (d) 7% compounded monthly: $ Round your answers to the nearest cent.
Mathematics
1 answer:
Scrat [10]2 years ago
7 0

The interest is a) $7000

b) $17709.73

c) $18672.62

d) $18901.67

What is the formula for simple and compound interest?

Simple interest = (P× r× t)

Compound interest = P(1+r/n)^nt - P

We will find the interest as shown below:

P=$4,000

t=25 years

a) r=7%=0.07

Simple interest = (P× r× t)

= (4000×0.07×25)

= $7000

b) r=7%=0.07

Compound interest = P(1+r)^t - P

= 4000(1+0.07)^25-4000

= $17709.73056

rounding to nearest cents

= $17709.73

c) r=7%=0.07

n=4

Compound interest = P(1+r/n)^nt - P

= 4000(1+0.07/4)^(25*4)-4000

= $18672.62375

rounding to nearest cent

= $18672.62

d) r=7%=0.07

n=12

Compound interest = P(1+r/n)^nt - P

= 4000(1+0.07/12)^(25*12)-4000

= $18901.6728

rounding to nearest cent

= $18901.67

Hence, the interest is a) $7000

b) $17709.73

c) $18672.62

d) $18901.67

Learn more about Interest here:

brainly.com/question/21104975

#SPJ1

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2 years ago
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zysi [14]

Given: The following functions

A)cos^2\theta=sin^2\theta-1B)sin\theta=\frac{1}{csc\theta}\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}

B

\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}

C

\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}

D

\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}

E

\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}

Hence, the following are identities

\begin{gathered} B)sin\theta=\frac{1}{csc\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}

The marked are the trigonometric identities

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