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Sav [38]
3 years ago
10

Richard’s checking account balance was $57.34 at the beginning of the week. During the week, he recorded the transactions below.

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
5 0

Richard’s account balance at the end of the week is $ 132.45

<h3>What is Account Balance ?</h3>

The amount of money in a person's bank account is called the account balance of that person

It is given that

Account balance of Richard’s at the beginning of the week = $57.34

Deposits in the week  : $ 163.75

Expenses in the week : Groceries + Credit card bill + Gas

Total Expenses in the week = 25.37 + 50 + 13.27

                                               = $ 88.64

The Richard’s account balance at the end of the week = Account Balance +Total Deposits - Total Withdrawals

= 57.34 + 163.75 - 88.64

Therefore , Richard’s account balance at the end of the week is $ 132.45.

To know more about Account Balance

brainly.com/question/23271078

#SPJ1

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Solve the inequality 7x+5&gt;2x+35
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You are riding along a straight river from east to the west at speed of 160m/m. At a given time, you will see the bearing of a c
yuradex [85]

Using the given information, the width of the river is 505.96 m

<h3>Bearing and Distances</h3>

From the question, we are to determine the width of the river

Consider the given diagram,

Let F be the first position of the biker

S be the second position

and C be the position of the church

First, we will determine the distance covered from F to S

Speed = 160m/min

Time = 4 minutes

Using the formula,

Distance = Speed × Time

Distance covered = 160 × 4

Distance covered = 640 m

That is, /FS/ = 640 m

Now, consider ΔFSC

∠F = 90° - 70° = 20°

∠S = 90° + 56° = 146°

and ∠C = 180° - 20° - 146° = 14°

By the Law of Sines

\frac{/FC/}{sin\ S}= \frac{/FS/}{sin\ C}

∴ \frac{/FC/}{sin\ 146^\circ}= \frac{640}{sin\ 14^\circ}

/FC/ =\frac{640 \times sin \ 146^\circ}{sin \ 14^\circ}

/FC/ = 1479.33 m

Now,

By SOH CAH TOA

cos \ 70^\circ = \frac{d}{/FC/}

NOTE: d is the width of the river

∴ d = /FC/ × cos70°

d = 1479.33 × cos70°

d = 505.96 m

Hence, the width of the river is 505.96 m

Learn more on Bearing and distances here: brainly.com/question/19351991

#SPJ1

7 0
2 years ago
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