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Zinaida [17]
3 years ago
6

The Seminole corporation has income before tax of $150,000.00 and taxable income of $65,000.00. The income tax rate is 35 percen

t. The journal entry includes a debit to income tax expense for what amount?
Mathematics
2 answers:
icang [17]3 years ago
6 0

Answer:

$29,750

Step-by-step explanation:

We have been given the income before tax: $150,000

And taxable income: $65,000

So, the income tax would be : $150,000-$65,000=$85000

So, 35% of $85000 is 85000\cdot \frac{35}{100}

On simplification we get:  $29,750

Hence, the income tax expense will be $29,750


cluponka [151]3 years ago
6 0

Financial Operations of Corporations

1. $17,000.00

2. $52,500.00

3. return on equity ratio

4. $2.37

5. no journal entry is required

6. $1,000.00

7. depreciation

8. retained earnings

9. $25,000.00

10. income tax payable

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Given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle
Anastasy [175]

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Part 1) False

Part 2) False

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

(x-h)^{2} +(y-k)^{2}=r^{2}

where

(h,k) is the center and r is the radius

In this problem the distance between the center and a point on the circle is equal to the radius

The formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

Part 1) given the center of the circle (-3,4) and a point on the circle (-6,2), (10,4) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

(x+3)^{2} +(y-4)^{2}=r^{2}

Find the distance (radius) between the center (-3,4) and (-6,2)

substitute in the formula of distance

r=\sqrt{(2-4)^{2}+(-6+3)^{2}}

r=\sqrt{(-2)^{2}+(-3)^{2}}

r=\sqrt{13}\ units

The equation of the circle is equal to

(x+3)^{2} +(y-4)^{2}=(\sqrt{13}){2}

(x+3)^{2} +(y-4)^{2}=13

Verify if the point (10,4) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=10,y=4

substitute

(10+3)^{2} +(4-4)^{2}=13

(13)^{2} +(0)^{2}=13

169=13 -----> is not true

therefore

The point is not on the circle

The statement is false

Part 2) given the center of the circle (1,3) and a point on the circle (2,6), (11,5) is on the circle.

true or false

substitute the center of the circle in the equation in standard form

(x-1)^{2} +(y-3)^{2}=r^{2}

Find the distance (radius) between the center (1,3) and (2,6)

substitute in the formula of distance

r=\sqrt{(6-3)^{2}+(2-1)^{2}}

r=\sqrt{(3)^{2}+(1)^{2}}

r=\sqrt{10}\ units

The equation of the circle is equal to

(x-1)^{2} +(y-3)^{2}=(\sqrt{10}){2}

(x-1)^{2} +(y-3)^{2}=10

Verify if the point (11,5) is on the circle

we know that

If a ordered pair is on the circle, then the ordered pair must satisfy the equation of the circle

For x=11,y=5

substitute

(11-1)^{2} +(5-3)^{2}=10

(10)^{2} +(2)^{2}=10

104=10 -----> is not true

therefore

The point is not on the circle

The statement is false

7 0
3 years ago
Please help due soon I don’t get it
natka813 [3]

Answer:

x>0 and y>0

Step-by-step explanation:

1= ++ 2= -+ 3= -- 4= +-

6 0
3 years ago
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