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Vesna [10]
3 years ago
9

As of December 31, 2020 $ 13,000. As of December 31, 2021 $ 25,000The enacted tax rate was 25% for 2020 and thereafter.What shou

ld be the balance in Kent's deferred tax liability account as of December 31, 2021?
Mathematics
1 answer:
blondinia [14]3 years ago
8 0

Answer:

$3,000

Step-by-step explanation:

The computation of the balance in Kent deferred tax liability is shown below:        

= {(December 31 2021) × (enacted tax rate)} - {(December 31 2020) × (Enacted tax rate)}

= {($25,000 × 25%) - ($13,000 × 25%)}

= $6,250 - $3,250

= $3,000

Along with the journal entry is also shown for better understanding

Income tax expense $41,500

       To Deferred tax liability $3,000

       To Income tax payable $38,250    ($153,000 × 25%)

(Being the income tax expense is recorded)

We debited the income tax expense as it increases the expenses account and on other side we credited the other two accounts as these are also increased

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Four times the cube of a number is less than the difference between forty and the number. Select the inequality that represents
tigry1 [53]

<u>Answer:</u>

4x^3 < 40 - x

<u>Explanation:</u>

let the number be x, now we are given that four times the cube of a number which means four multiplied by the cube of number

i.e 4 multiplied by x^3 is less than the difference of 40 and the number x i.e 40 - x

x is for sure less than 40 as x^3 is less than the difference between 40 and x so if we write x - 40 then we will obtain negative value which will be greater than x^3.

So, the inequality obtained can be written as 4x^3 < 40 - x

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3 years ago
Read 2 more answers
PLEASE SHOW ALL THE STEPS THAT YOU USE TO SOLVE THIS PROBLEM
Mademuasel [1]

Answer:

{x = 1 , y=1, z=0

Step-by-step explanation:

Solve the following system:

{-2 x + 2 y + 3 z = 0 | (equation 1)

{-2 x - y + z = -3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x - 3 y - 2 z = -3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+3 y + 2 z = 3 | (equation 2)

{2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+3 y + 2 z = 3 | (equation 2)

{0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y + 6 z = 5 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+5 y+0 z = 5 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Divide equation 2 by 5:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:

{-(2 x) + 0 y+3 z = -2 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Subtract 3 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = -2 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = 1 | (equation 1)

{0 x+y+0 z = 1 | (equation 2)

{0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer:  {x = 1 , y=1, z=0

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Answer:

the Answer will be letter B

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Need to keep cell bill under 60.,monthly charge is 30. And texts are 0.05 each. Write an inequality to represent the # of texts
nignag [31]

Answer:

30 + 0.05x < 60

The number of texts she can have is less than 600.

Step-by-step explanation:

Total monthly charge

30 + 0.05 for each text.

So, for x tests, the charge is:

C(x) = 30 + 0.05x

Write an inequality to represent the # of texts she can have?

It has to be under 60, so:

C(x) < 60

30 + 0.05x < 60

The solution is:

0.05x < 30

x < \frac{30}{0.05}

x < 600

The number of texts she can have is less than 600.

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What is the answer to 72% of 14?
SSSSS [86.1K]
14/100 = 0.14
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