Which value is the solution to the open sentence? 2 + 3x= 5

Feliz [49]

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

1. Allowance for Doubtful Accounts

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

Accounts Receivable

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%)

Allowance for Doubtful Accounts

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

The number of students in sixth grade is 1.03 times the number of students in seventh grade. What percent is equivalent to 1.03?

Minchanka [31]

Move the decimal to the right twice to make this into a percent:

1.03 —-> 103%

8 0

2 years ago

Select all expressions that represent a correct solution to the equation 6(x+4)=20

Sergeeva-Olga [200]

Answer:

D. 20 ÷ 6 − 4

E. 1/6(20 − 24)

F. (20−24) ÷ 6

Step-by-step explanation:

6(x + 4) = 20

6(x) + 6(4) = 20

6x + 24 = 20

- 24 - 24

6x = -4

/6 /6

x = -4/6 or x = -2/3

D. 20 ÷ 6 − 4

20/6 - 4

20/6 - 24/6

-4/6 = -2/3

E. 1/6(20 − 24)

1/6(-4)

-4/6 = -2/3

F. (20 − 24) ÷ 6

- 4 / 6

-4/6 = -2/3

Hope this helps!

3 0

2 years ago

This math my graduation depends on it

In-s [12.5K]

In an arithmetic sequence, consecutive terms have a fixed distance d between them. If a₁ is the first term, then

2nd term = a₂ = a₁ + d

3rd term = a₃ = a₂ + d = a₁ + 2d

4th term = a₄ = a₃ + d = a₁ + 3d

and so on, up to

nth term = $a_n = a_{n-1} + d = a_{n-2} + 2d = a_{n-3} + 3d = \cdots = a_1 + (n-1)d$

so that every term in the sequence can be expressed in terms of a₁ and d.

6. It's kind of hard to tell, but it looks like you're given a₁₃ = -53 and a₃₅ = -163.

We have

a₁₃ = a₁ + 12d = -53

a₃₅ = a₁ + 34d = -163

Solve for a₁ and d. Eliminating a₁ and solving for d gives

(a₁ + 12d) - (a₁ + 34d) = -53 - (-163)

-22d = 110

d = -5

and solving for a₁, we get

a₁ + 12•(-5) = -53

a₁ - 60 = -53

a₁ = 7

Then the nth term is recursively given by

$a_n = a_{n-1}-5$

and explicitly by

$a_n = 7 + (n-1)(-5) = 12 - 5n$

7. We do the same thing here. Use the known terms to find a₁ and d :

a₁₉ = a₁ + 18d = 15

a₃₈ = a₁ + 37d = 72

⇒ (a₁ + 18d) - (a₁ + 37d) = 15 - 72

⇒ -19d = -57

⇒ d = 3

⇒ a₁ + 18•3 = 15

⇒ a₁ = -39

Then the nth term is recursively obtained by

$a_n = a_{n-1}+3$

and explicitly by

$a_n = -39 + (n-1)\cdot3 = 3n-42$

8. I won't both reproducing the info I included in my answer to your other question about geometric sequences.

We're given that the 1st term is 3 and the 2nd term is 12, so the ratio is r = 12/3 = 4.

Then the next three terms in the sequence are

192 • 4 = 768

768 • 4 = 3072

3072 • 4 = 12,288

The recursive rule with a₁ = 3 and r = 4 is

$a_n = 4a_{n-1}$

and the explicit rule would be

$a_n = 3\cdot4^{n-1}$

7 0

2 years ago

Triangle EFG is dilated by a scale factor of one half centered at (1, 1) to create triangle E'F'G'. Which statement is true abou

olganol [36]

Answer:

A triangle is dilated by scale factor of 1/5

Step-by-step explanation:

A triangle has three sides. All the corresponding angles of triangle must be congruent. When the triangle is dilated it is rotated 90 degrees clockwise about its origin. The shape of the triangle will not change due to rotation of dilation.

6 0

3 years ago