Test your conjectures. Use the reverse tabular method to find the quotient. (3x^5- 2x^4+6x^3- 4x^2- 24x +16)/(x^2+4)

Mr. and Mrs. Petersen filed a joint tax return in 2015. The couple divorced in 2016.The IRS audited their 2015 return during 201

Harlamova29_29 [7]

Answer:

Step-by-step explanation:

If taxes have been underpaid, it won’t matter to the IRS who is responsible. If the return was filed jointly, the government can go after both even if one didn’t personally earn one penny of the reported income.

Hence in the given case,

For the unpaid amount of $38,200 IRS can assess either Mr.peterson or Mrs.Peterson for the entire deficiency as both are jointly and severally liable for the tax paid before they got divorced.

6 0

2 years ago

Whats the gcf of 15 and 70

makvit [3.9K]

The gcf of 15 and 70 is 5

8 0

3 years ago

Read 2 more answers

Analyze the diagram below and complete the instructions that follow. 550 450

svetlana [45]

Answer:

130 degrees

Step-by-step explanation:

let the angle be x

2x+45+55=360(Sum of angles in quad)

2x=260

x=130

6 0

3 years ago

The triangles are similar. What is the value of x? x =

saveliy_v [14]

Answer:

x = 23

Step-by-step explanation:

Since the triangles are similar then the ratios of corresponding sides are equal, that is

$\frac{39}{x-10}$ = $\frac{36}{12}$ = $\frac{15}{5}$ = 3

multiply both sides by (x - 10)

3(x - 10) = 39 ( divide both sides by 3 )

x - 10 = 13 ( add 10 to both sides )

x = 23

8 0

3 years ago

Read 2 more answers

1. The height of a triangle is 6 m more than its base. The area of the triangle is 56 m². What is the length of the base? Enter

Elodia [21]

Answers:
1. 8 m
2. 17 m
3. 7 cm
4. 2 s

Explanations:

1. Let x = length of the base
x + 6 = height of the base

Then, the area of the triangle is given by

(Area) = (1/2)(base)(height)
56 = (1/2)(x)(x + 6)
56 = (1/2)(x²  + 6x)

Using the symmetric property of equations, we can interchange both sides of equations so that

(1/2)(x²  + 6x) = 56

Multiplying both sides by 2, we have

x² + 6x = 112

The right side should be 0. So, by subtracting both sides by 112, we have

x² + 6x - 112 = 112 - 112
x² + 6x - 112 = 0

By factoring, x² + 6x - 112 = (x - 8)(x + 14). So, the previous equation becomes

(x - 8)(x +14) = 0

So, either

x - 8 = 0 or x + 14 = 0

Thus, x = 8 or x = -14. However, since x represents the length of the base and the length is always positive, it cannot be negative. Hence, x = 8. Therefore, the length of the base is 8 cm.

2. Let x = length of increase in both length and width of the rectangular garden

Then,

14 + x = length of the new rectangular garden
12 + x = width of the new rectangular garden

So,

(Area of the new garden) = (length of the new garden)(width of the new garden)

255 = (14 + x)(12 + x) (1)

Note that

(14 + x)(12 + x) = (x + 14)(x + 12)
= x(x + 14) + 12(x + 14)
= x² + 14x + 12x + 168
= x² + 26x + 168

So, the equation (1) becomes

255 = x² + 26x + 168

By symmetric property of equations, we can interchange the side of the previous equation so that

x² + 26x + 168 = 255

To make the right side becomes 0, we subtract both sides by 255:

x² + 26x + 168 - 255 = 255 - 255
x² + 26x - 87 = 0

To solve the preceding equation, we use the quadratic formula.

First, we let

a = numerical coefficient of x² = 1

Note: if the numerical coefficient is hidden, it is automatically = 1.

b = numerical coefficient of x = 26
c = constant term = - 87

Then, using the quadratic formula

$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} = \frac{-26 \pm \sqrt{26^2 - 4(1)(-87)} }{2(1)} 
\newline x = \frac{-26 \pm \sqrt{1,024} }{2}
\newline
\newline x = \frac{-26 \pm 32 }{2}$

So,

$x = \frac{-26 + 32 }{2} \text{ or } x = \frac{-26 - 32 }{2}
\newline x = \frac{6 }{2} \text{ or } x = \frac{-58 }{2}
\newline \boxed{ x = 3 \text{ or } x = -29}$

Since x represents the amount of increase, x should be positive.

Hence x = 3.

Therefore, the length of the new garden is given by

14 + x = 14 + 3 = 17 m.

3. The area of the shaded region is given by

(Area of shaded region) = π(outer radius)² - π(inner radius)²
= π(2x)² - π6²
= π(4x² - 36)

Since the area of the shaded region is 160π square centimeters,

π(4x² - 36) = 160π

Dividing both sides by π, we have

4x² - 36 = 160

Note that this equation involves only x² and constants. In these types of equation we get rid of the constant term so that one side of the equation involves only x² so that we can solve the equation by getting the square root of both sides of the equation.

Adding both sides of the equation by 36, we have

4x² - 36 + 36 = 160 + 36
4x² = 196

Then, we divide both sides by 4 so that

x² = 49

Taking the square root of both sides, we have

$x = \pm 7$

Note: If we take the square root of both sides, we need to add the plus minus sign $(\pm)$ because equations involving x² always have 2 solutions.

So, x = 7 or x = -7.

But, x cannot be -7 because 2x represents the length of the outer radius and so x should be positive.

Hence x = 7 cm

4. At time t, h(t) represents the height of the object when it hits the ground. When the object hits the ground, its height is 0. So,

h(t) = 0 (1)

Moreover, since $v_0 = 27$ and $h_0 = 10$ ,

h(t) = -16t² + 27t + 10 (2)

Since the right side of the equations (1) and (2) are both equal to h(t), we can have

-16t² + 27t + 10 = 0

To solve this equation, we'll use the quadratic formula.

Note: If the right side of a quadratic equation is hard to factor into binomials, it is practical to solve the equation by quadratic formula.

First, we let

a = numerical coefficient of t² = -16
b = numerical coefficient of t = 27
c = constant term = 10

Then, using the quadratic formula

$t = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} = \frac{-27 \pm \sqrt{27^2 - 4(-16)(10)} }{2(-16)} \newline t = \frac{-27 \pm \sqrt{1,369} }{-32} \newline \newline t = \frac{-27 \pm 37 }{32}$

So,

$t = \frac{-27 + 37 }{-32} \text{ or } t = \frac{-27 - 37 }{-32} \newline t = \frac{-10}{32} \text{ or } t = \frac{-64 }{-32} \newline \boxed{ t = -0.3125 \text{ or } t = 2}$

Since t represents the amount of time, t should be positive.

Hence t = 2. Therefore, it takes 2 seconds for the object to hit the ground.

3 0

3 years ago

Read 2 more answers