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3 years ago

Chadwick contributes 17% of his $68,600 annual salary to his 401(k) plan. What is his pre-tax income?

Mathematics

2 answers:

dangina [55]3 years ago

4 0

Answer:

56938

Step-by-step explanation:

Tanzania [10]3 years ago

4 0

Answer:

His pre-tax income is $56938.

Step-by-step explanation:

Consider the provided statement:

Chadwick contributes 17% of his annual salary to his 401(k) plan.

Therefore,

$\frac{17}{100} $68,600=17($686)$

$=11662$

Thus, Chadwick contributes $11662 in 401(k) plan.

Pretax earnings exclude taxes, therefore subtract $11662 from his annual salary.

$68600-$11662=$56938

Hence, his pre-tax income is $56938.

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On Monday, 263 fourth graders went to the Perot Museum with their class. On Tuesday, 187 fifth graders went. Each student paid $

ira [324]

Answer:

About $35

Step-by-step explanation:

3 0

3 years ago

The first term of an arithmetic sequence is -3 and the fifteenth term is 53. What is the common difference of the sequence?

Maru [420]

C: 4 is the right answer

Step-by-step explanation:

Given

a1 = -3

a15 = 53

We know that explicit formula for the arithmetic sequence is:

$a_n=a_1+(n-1)d$

For the 15th, term it will be

$a_{15}=-3+(15-1)d\\53=-3+14d$

Adding 3 on both sides

$53+3 = -3+3 + 14d\\56 = 14d$

Dividing both sides by 14

$\frac{56}{14}=\frac{14d}{14}\\d=4$

Hence,

C: 4 is the right answer

Keywords: Arithmetic sequence, Common Difference

Learn more about arithmetic sequence at:

#LearnwithBrainly

4 0

3 years ago

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2

Alika [10]

As the ladder is pulled away from the wall, the area and the height with the

wall are decreasing while the angle formed with the wall increases.

The correct response are;

(a) The velocity of the top of the ladder = 1.5 m/s downwards

(b) The rate the area formed by the ladder is changing is approximately -75.29 ft.²/sec

Reasons:

The given parameter are;

Length of the ladder, l = 25 feet

Rate at which the base of the ladder is pulled, $\displaystyle \frac{dx}{dt}$ = 2 feet per second

(a) Let y represent the height of the ladder on the wall, by chain rule of differentiation, we have;

$\displaystyle \frac{dy}{dt} = \mathbf{\frac{dy}{dx} \times \frac{dx}{dt}}$

25² = x² + y²

y = √(25² - x²)

$\displaystyle \frac{dy}{dx} = \frac{d}{dx} \sqrt{25^2 - x^2} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}$

Which gives;

$\displaystyle \frac{dy}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times \frac{dx}{dt} = \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2$

$\displaystyle \frac{dy}{dt} = \mathbf{ \frac{x \cdot \sqrt{625-x^2} }{x^2- 625}\times2}$

When x = 15, we get;

$\displaystyle \frac{dy}{dt} = \frac{15 \times \sqrt{625-15^2} }{15^2- 625}\times2 = \mathbf{-1.5}$

The velocity of the top of the ladder = 1.5 m/s downwards

When x = 20, we get;

$\displaystyle \frac{dy}{dt} = \frac{20 \times \sqrt{625-20^2} }{20^2- 625}\times2 = -\frac{8}{3} = -2.\overline 6$

The velocity of the top of the ladder = $\underline{-2.\overline{6} \ m/s \ downwards}$

When x = 24, we get;

$\displaystyle \frac{dy}{dt} = \frac{24 \times \sqrt{625-24^2} }{24^2- 625}\times2 = \mathbf{-\frac{48}{7}} \approx -6.86$

The velocity of the top of the ladder ≈ -6.86 m/s downwards

(b) $\displaystyle The \ area\ of \ the \ triangle, \ A =\mathbf{\frac{1}{2} \cdot x \cdot y}$

Therefore;

$\displaystyle The \ area\ A =\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}$

$\displaystyle \frac{dA}{dx} = \frac{d}{dx} \left (\frac{1}{2} \cdot x \cdot \sqrt{25^2 - x^2}\right) = \mathbf{\frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250}}$

$\displaystyle \frac{dA}{dt} = \mathbf{ \frac{dA}{dx} \times \frac{dx}{dt}}$

Therefore;

$\displaystyle \frac{dA}{dt} = \frac{(2 \cdot x^2- 625)\cdot \sqrt{625-x^2} }{2\cdot x^2 - 1250} \times 2$

When the ladder is 24 feet from the wall, we have;

x = 24

$\displaystyle \frac{dA}{dt} = \frac{(2 \times 24^2- 625)\cdot \sqrt{625-24^2} }{2\times 24^2 - 1250} \times 2 \approx \mathbf{ -75.29}$

The rate the area formed by the ladder is changing, $\displaystyle \frac{dA}{dt}$ ≈ -75.29 ft.²/sec

$\displaystyle sin(\theta) = \frac{x}{25}$

$\displaystyle \theta = \mathbf{arcsin \left(\frac{x}{25} \right)}$

$\displaystyle \frac{d \theta}{dt} = \frac{d \theta}{dx} \times \frac{dx}{dt}$

$\displaystyle\frac{d \theta}{dx} = \frac{d}{dx} \left(arcsin \left(\frac{x}{25} \right) \right) = \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625}}$

Which gives;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-x^2} }{x^2 - 625}\times \frac{dx}{dt}= \mathbf{ -\frac{\sqrt{625-x^2} }{x^2 - 625} \times 2}$

When x = 24 feet, we have;

$\displaystyle \frac{d \theta}{dt} = -\frac{\sqrt{625-24^2} }{24^2 - 625} \times 2 \approx \mathbf{ 0.286}$

Rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 24 feet from the wall is $\displaystyle \frac{d \theta}{dt}$ ≈ 0.286 rad/sec