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Ugo [173]
3 years ago
8

During a phone call, Beyunka was told of the most recent transactions in her company's business account. There were deposits of

$1, 250 and $3,040.57, three withdrawals of $400 each, and the bank removed two separate $35 penalties to the account that resulted from the bank's errors. Based on the information, how much did the balance of the account change?
Mathematics
1 answer:
Norma-Jean [14]3 years ago
4 0

Answer:

The amount of change in the balance of the account is an increase of $3160.57.

Step-by-step explanation:

i) first deposit is given as $1250

ii) second deposit is given as $3040.57

iii) first withdrawal is given as $400

iv) second withdrawal is given as $400

v) third withdrawal is given as $400

vi) first penalty removed is $35

vii) second penalty removed is $35

viii) therefore the change to the balance is given by

$1250 + $3040.57 - $400 - $400 - $400 + $35 + $35 = $3160.57

viii) Therefore the amount of change in the balance of the account is an increase of $3160.57.

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serg [7]

Answer:

The result of the integral is:

\arcsin{(\frac{x}{3})} + C

Step-by-step explanation:

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We have the term in the following format: a^2 - x^2, in which a = 3.

In this case, the substitution is given by:

x = a\sin{\theta}

So

dx = a\cos{\theta}d\theta

In this question:

a = 3

x = 3\sin{\theta}

dx = 3\cos{\theta}d\theta

So

\int \frac{3\cos{\theta}d\theta}{\sqrt{9-(3\sin{\theta})^2}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9 - 9\sin^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{\theta})}}

We have the following trigonometric identity:

\sin^{2}{\theta} + \cos^{2}{\theta} = 1

So

1 - \sin^{2}{\theta} = \cos^{2}{\theta}

Replacing into the integral:

\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C

Coming back to x:

We have that:

x = 3\sin{\theta}

So

\sin{\theta} = \frac{x}{3}

Applying the arcsine(inverse sine) function to both sides, we get that:

\theta = \arcsin{(\frac{x}{3})}

The result of the integral is:

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Because, we know that \sec \theta =\frac{1}{\cos \theta} and \tan \theta = \frac{\sin \theta}{\cos \theta}

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