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weeeeeb [17]
3 years ago
11

Elijah waited until June u,2017 to file his 1040 for tax year 2016. His balance due is $88. What is the failure to file penalty?

Mathematics
1 answer:
Ad libitum [116K]3 years ago
7 0
The failure to file penalty refers to the amount of money that a tax payer has to pay as penalty for not filing his tax before the assigned deadline. The failure to pay penalty is 0.5% of the tax balance that is owned per month.
Thus, for Elijah, who is owning $88, the failure to pay penalty is 0.5% *88 = $0.44.
Elijah has to pay $0.44 per month for the period during which he withhold the tax.
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1. A hotel manager buys 30 pillow cases and 25 shower curtains for $375. The manager then buys another 28 pillow cases and 35 sh
skad [1K]

Answer:

For #2: It's C. 17 nickels and 15 quarters.

Step-by-step explanation:

17*5= 85

15*25= 375

375+85= 460

add in the decimal:

17* 5 cents= $.85

15* 25 cents= $3.75

$3.75+ $.85= $4.60

7 0
3 years ago
Find the vertex of the function given below.<br> y = 3x^2+ 6x+1
statuscvo [17]

Answer:

(-1,-2)

Step-by-step explanation:

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3 years ago
The ratio of boys to girls in the class is 4 to 5. If there are 18 boys and girls in the
bearhunter [10]

Answer:

8 boys and 10 girls

Step-by-step explanation:

If the ratio is 4:5,

4x2 + 5x2 equal 8+10 =

18.

4:5=8:10

7 0
3 years ago
Read 2 more answers
The expression x2y - 2xy - 24y can be factored by first factoring out a common factor of y. After the common factor is removed,
svp [43]

The remaining factor of x^2y - 2xy - 24y is (x - 6)(x + 4)

<h3>How to determine the remaining factor?</h3>

The expression is given as:

x^2y - 2xy - 24y

Factor out y from the expression

y(x^2 - 2x - 24)

Expand the equation

y(x^2 + 4x - 6x - 24)

Factorize

y(x - 6)(x + 4)

Hence, the remaining factor of x^2y - 2xy - 24y is (x - 6)(x + 4)

Read more about factored expression at:

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3 0
2 years ago
The line 5x – 5y = 2 intersects the curve x2y – 5x + y + 2 = 0 at
inna [77]

Answer:

(a) The coordinates of the points of intersection are (-2, -12/5), (2/5, 0), and (2, 8/5)

(b) The gradient of the curve at each point of intersection are;

Gradient at (-2, -12/5) = -0.92

Gradient at (2/5, 0) = 4.3

Gradient at (2, 8/5) = -0.28

Step-by-step explanation:

The equations of the lines are;

5·x - 5·y = 2......(1)

x²·y - 5·x + y + 2 = 0.......(2)

Making y the subject of equation (1) gives;

5·y = 5·x - 2

y = (5·x - 2)/5

Making y the subject of equation (2) gives;

y·(x² + 1) - 5·x + 2 = 0

y = (5·x - 2)/(x² + 1)

Therefore, at the point the two lines intersect their coordinates are equal thus we have;

y = (5·x - 2)/5 = y = (5·x - 2)/(x² + 1)

Which gives;

\dfrac{5 \cdot x - 2}{5} = \dfrac{5 \cdot x - 2}{x^2 + 1}

Therefore, 5 = x² + 1

x² = 5 - 1 = 4

x = √4 = 2

Which is an indication that the x-coordinate is equal to 2

The y-coordinate is therefore;

y = (5·x - 2)/5 = (5 × 2 - 2)/5 = 8/5

The coordinates of the points of intersection = (2, 8/5}

Cross multiplying the following equation

Substituting the value for y in equation (2) with (5·x - 2)/5 gives;

\dfrac{5 \cdot x^3 - 2 \cdot x^2 - 20 \cdot x + 8}{5} = 0

Therefore;

5·x³ - 2·x² - 20·x + 8 = 0

(x - 2)×(5·x² - b·x + c) = 5·x³ - 2·x² - 20·x + 8

Therefore, we have;

x²·b - 2·x·b -x·c + 2·c -5·x³ + 10·x²

5·x³ - 10·x² - x²·b + 2·x·b + x·c - 2·c = 5·x³ - 2·x² - 20·x + 8

∴ c = 8/(-2) = -4

2·b + c = - 20

b = -16/2 = -8

Therefore;

(x - 2)×(5·x² - b·x + c) = (x - 2)×(5·x² + 8·x - 4)

(x - 2)×(5·x² + 8·x - 4) = 0

5·x² + 8·x - 4 = 0

x² + 8/5·x - 4/5  = 0

(x + 4/5)² - (4/5)² - 4/5 = 0

(x + 4/5)² = 36/25

x + 4/5 = ±6/5

x = 6/5 - 4/5 = 2/5 or -6/5 - 4/5 = -2

Hence the three x-coordinates are

x = 2, x = - 2, and x = 2/5

The y-coordinates are derived from y = (5·x - 2)/5 as y = 8/5, y = -12/5, and y = y = 0

The coordinates of the points of intersection are (-2, -12/5), (2/5, 0), and (2, 8/5)

(b) The gradient of the curve, \dfrac{\mathrm{d} y}{\mathrm{d} x}, is given by the differentiation of the equation of the curve, x²·y - 5·x + y + 2 = 0 which is the same as y = (5·x - 2)/(x² + 1)

Therefore, we have;

\dfrac{\mathrm{d} y}{\mathrm{d} x}= \dfrac{\mathrm{d} \left (\dfrac{5 \cdot x - 2}{x^2 + 1}  \right )}{\mathrm{d} x} = \dfrac{5\cdot \left ( x^{2} +1\right )-\left ( 5\cdot x-2 \right )\cdot 2\cdot x}{\left (x^2 + 1 ^{2} \right )}.......(3)

Which gives by plugging in the value of x in the slope equation;

At x = -2, \dfrac{\mathrm{d} y}{\mathrm{d} x} = -0.92

At x = 2/5, \dfrac{\mathrm{d} y}{\mathrm{d} x} = 4.3

At x = 2, \dfrac{\mathrm{d} y}{\mathrm{d} x} = -0.28

Therefore;

Gradient at (-2, -12/5) = -0.92

Gradient at (2/5, 0) = 4.3

Gradient at (2, 8/5) = -0.28.

7 0
3 years ago
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