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Maksim231197 [3]
3 years ago
15

The area of a quadrilateral is 64 in^2. The length of the quadrilateral is 32 in. How wide is the quadrilateral?

Mathematics
1 answer:
Mars2501 [29]3 years ago
7 0

Answer: 2 inches

Explanation: The area formula is length times width (L*w=a). We know that the area is 64in^2 (a=64) and the length is 32in (L=32), so we put those values into the equation and get 32*w=64. We then get w by itself by dividing both sides by 32, and get w=2. so the width is 2 inches.

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Allowance method entries
Feliz [49]

Using the Allowance Method, the relevant transactions can be completed in the books of Wild Trout Gallery as follows:

1. <u>Allowance for Doubtful Accounts</u>

Accounts                                          Debit       Credit

Jan. 1 Beginning balance                             $53,800

Jan. 19 Accounts Receivable                           2,560

Apr. 3 Accounts Receivable       $14,670

July 16 Accounts Receivable        19,725

Nov. 23 Accounts Receivable                         4,175

Dec. 31 Accounts Receivable       25,110

Dec. 31 Ending balance          $56,500

Dec. 31 Bad Debts Expenses                   $55,470

Totals                                        $116,005  $116,005

<u>Accounts Receivable</u>

Accounts                                          Debit               Credit

Jan. 1 Beginning balance           $2,290,000

Jan. 19 Allowance for Doubtful           2,560

Jan. 19 Cash                                                            $2,560

Apr. 3  Allowance for Doubtful                                14,670

July 16  Allowance for Doubtful                              19,725

July 16  Cash                                                             6,575

Nov. 23  Allowance for Doubtful         4,175

Nov. 23 Cash                                                             4,175

Dec. 31  Allowance for Doubtful                             25,110

Dec. 31   Sales Revenue            8,020,000

Dec. 31   Cash                                               $8,944,420

Dec. 31 Ending balance                                 $1,299,500

Totals                                        $10,316,735 $10,316,735

3. Expected net realizable value of the accounts receivable as of December 31 = $1,243,000 ($1,299,500 - $56,500)

Allowance for Doubtful Accounts ending balance = $40,100 ($8,020,000 x 0.5%)

<u>Allowance for Doubtful Accounts</u>

Accounts                                          Debit       Credit

Jan. 1 Beginning balance                             $53,800

Jan. 19 Accounts Receivable                           2,560

Apr. 3 Accounts Receivable       $14,670

July 16 Accounts Receivable        19,725

Nov. 23 Accounts Receivable                         4,175

Dec. 31 Accounts Receivable       25,110

Dec. 31 Ending balance           $40,100

Dec. 31 Bad Debts Expenses                  $39,070

Totals                                        $99,605   $99,605

4. a. Bad Debt Expense for the year = $39,070

4.b. Balance for Allowance Accounts = $40,100

4.c. Expected net realizable value of the accounts receivable = $1,259,400 ($1,299,500 - $40,100)

Data Analysis:

Jan. 19 Accounts Receivable $2,560 Allowance for Uncollectible Accounts $2,560

Jan. 19 Cash $2,560 Accounts Receivable $2,560

Apr. 3 Allowance for Uncollectible Accounts $14,670 Accounts Receivable $14,670

July 16 Cash $6,575 Allowance for Uncollectible Accounts $19,725 Accounts Receivable $26,300

Nov. 23 Accounts Receivable $4,175 Allowance for Uncollectible Accounts $4,175

Nov. 23 Cash $4,175 Accounts Receivable $4,175

Dec. 31 Allowance for Uncollectible Accounts $25,110 Accounts Receivable $25,110

Accounts Receivable ending balance = $1,299,500

Allowance for Uncollectible Accounts ending balance = $56,500

Learn more: brainly.com/question/22984282

4 0
3 years ago
Find the radius of a sphere with a volume of 4.5π<br> cubic centimeters.
garik1379 [7]
R≈1.02 
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8 0
3 years ago
A spherical water tank has a diameter of 14.5 feet.
lawyer [7]

Answer:

1595.45 ft3

Step-by-step explanation:

The formula is (4/3) *3.14*r^3

since the diameter is 14.5, the radius is 7.25

(4/3) *3.14*(7.25^3)

7 0
4 years ago
Simplify - tanx-(sec^2 / tanx)
Mashcka [7]

Answer:

-cot(x)

Step-by-step explanation:

tanx-(sec^2x  / tanx)

Lets get a common denominator of tan

tan x * tan x/ tan  -sec ^2 x/ tan x

tan ^2 x - sec ^2 x

---------------------------

tan x

We know than sec^2 - tan^2 =1  (trig identity)  so factor out -1

-1(-tan ^2 x + sec ^2 x)

---------------------------

tan x

-1(1)

---------------------------

tan x

-1

----------

tan (x)

We know  1/ tan x = cot (x)

- cot(x)

7 0
3 years ago
It is given a polynomial f(x) = 2x³ + ax² - bx + 12, where a and b are constants. When f(x) is divided by x-2, the remainder is
VMariaS [17]

Answer:

Step-by-step explanation:

<em>(a).</em> 2( 2 )³ + a ( 2 )² - b ( 2 ) + 12 + ( 18 ) = 0

16 + 4a - 2b + 30 = 0

4a - 2b = - 46

<em>2a - b = - 23</em> ....... <em>( 1 )</em>

2 ( - 4 )³ + a ( - 4 )² - b ( - 4 ) + 12 = 0

- 128 + 16a + 4b + 12 = 0

16a + 4b = 116

<em>4a + b = 29</em> ........ <em>( 2 )</em>

<em>( 1 )</em> + <em>( 2 )</em>

6a = 6 , <em>a = 1</em>

4(1) + b = 29 , <em>b = 25</em>

<em>(b).</em> 2x³ + x² + 25x + 12 = 0

x_{1} = <em>- 4</em> ( given: f(x) is divisible by x + 4 )

x_{2} = <em>0.5</em>

x_{3} = <em>3</em>

5 0
3 years ago
Read 2 more answers
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