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ruslelena [56]
3 years ago
15

Last year, Justin opened an investment account with $5400. At the end of the year, the amount in the account had increased by 6.

5%. How much is this increase in dollars? How much money was in his account at the end of last year?
Mathematics
2 answers:
lara31 [8.8K]3 years ago
6 0

Answer:

$351 and $5751

Step-by-step explanation:

Given, Justin opened an investment account with $5400. At the end of the year, 6.5% increased. So, the total increase in the dollar is 5400* 0.065 = 351. The increase in the dollar is $364.

At the end of the year, Justin will have 5400 + 351 = $5,751. So, Justin will have $5,751 at the end of the year in his investment bank account.

maksim [4K]3 years ago
5 0

Answer:

$351

$5,751

Step-by-step explanation:

Justin opened an investment account with $5400.

At the end of the year, the amount in the account had increased by 6.5%.

So,

$5,400 - 100%

$x - 6.5%

Write a proportion:

\dfrac{5,400}{x}=\dfrac{100}{6.5}

Cross multiply:

100x=5,400\cdot 6.5\\ \\x=54\cdot 6.5\\ \\x=\$351

Justin's increase in dollars is $351.

In his account at the end of the year was

\$5,400+\$351=\$5,751

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Ludmilka [50]

Answer:

(a)\ \sec^2(\theta) = 82

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Step-by-step explanation:

Given

\tan(\theta) = 9

Required

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Using tan formula, we have:

\tan(\theta) = \frac{Opposite}{Adjacent}

This gives:

\frac{Opposite}{Adjacent} = 9

Rewrite as:

\frac{Opposite}{Adjacent} = \frac{9}{1}

Using a unit ratio;

Opposite = 9; Adjacent = 1

Using Pythagoras theorem, we have:

Hypotenuse^2 = Opposite^2 + Adjacent^2

Hypotenuse^2 = 9^2 + 1^2

Hypotenuse^2 = 81 + 1

Hypotenuse^2 = 82

Take square roots of both sides

Hypotenuse =\sqrt{82}

So, we have:

Opposite = 9; Adjacent = 1

Hypotenuse =\sqrt{82}

Solving (a):

\sec^2(\theta)

This is calculated as:

\sec^2(\theta) = (\sec(\theta))^2

\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2

Where:

\cos(\theta) = \frac{Adjacent}{Hypotenuse}

\cos(\theta) = \frac{1}{\sqrt{82}}

So:

\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2

\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2

\sec^2(\theta) = (\sqrt{82})^2

\sec^2(\theta) = 82

Solving (b):

\cot(\theta)

This is calculated as:

\cot(\theta) = \frac{1}{\tan(\theta)}

Where:

\tan(\theta) = 9 ---- given

So:

\cot(\theta) = \frac{1}{\tan(\theta)}

\cot(\theta) = \frac{1}{9}

Solving (c):

\cot(\frac{\pi}{2} - \theta)

In trigonometry:

\cot(\frac{\pi}{2} - \theta) = \tan(\theta)

Hence:

\cot(\frac{\pi}{2} - \theta) = 9

Solving (d):

\csc^2(\theta)

This is calculated as:

\csc^2(\theta) = (\csc(\theta))^2

\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2

Where:

\sin(\theta) = \frac{Opposite}{Hypotenuse}

\sin(\theta) = \frac{9}{\sqrt{82}}

So:

\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2

\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2

\csc^2(\theta) = \frac{82}{81}

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3 years ago
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<h3>What are fractions?</h3>

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Since there are 6 cups in the box, it means the fraction of the box of pancake that Eliza used is;

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ivolga24 [154]

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