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olga_2 [115]
3 years ago
5

In 2008, the Social Security tax percentage was 6.2% for the first $102,000 earned. The Medicare tax percentage was 1.45% of you

r entire salary. Darby made $143,000 in 2008. How much did Darby pay in Social Security and Medicare taxes that year?​
Mathematics
1 answer:
Shalnov [3]3 years ago
4 0

Answer:

In 2008, the Social Security percentage was 6.2% for the first $102,000 earned. The Medicare percentage was 1.45% of your entire salary. Darby made $143,000 in 2008.

mark as brainiest

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Terry Sheehan owns property assessed at $390,000 and pays a property tax rate of 1.69 per hundred dollars of assessed value. He
34kurt

Answer:

$1689

Step-by-step explanation:

Excess of rent amount over annual property tax bill = annual rent - annual property tax

annual property tax =  [(assessed property value / 100) x 1.69]

( $390,000 / 100) x 1.69 = $6591

Annual rent = $690 x 12 = $8280

$8280 - $6591 = $1689

7 0
3 years ago
Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
kherson [118]

Answer:

The answer is

f(x) = {\displaystyle  8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }

Step-by-step explanation:

Remember that Taylor says that

f(x) = {\displaystyle \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a) }{k!}(x-a)^k }

For this case

f^{(0)} (-2) = 8(-2)-3(-2)^3 = 8\\f^{(1)} (-2) = 8-3(3)(-2)^2 = -28\\f^{(2)} (-2) = -3(3)2(-2) = -36\\f^{(2)} (-2) = -3(3)2 = -18

f(x) = {\displaystyle  8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }

5 0
3 years ago
Find the area of the region enclosed by the graphs of the functions
Vaselesa [24]

Answer:

\displaystyle A = \frac{8}{21}

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Algebra I</u>

  • Terms/Coefficients
  • Functions
  • Function Notation
  • Graphing
  • Solving systems of equations

<u>Calculus</u>

Area - Integrals

Integration Rule [Reverse Power Rule]:                                                                 \displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C

Integration Rule [Fundamental Theorem of Calculus 1]:                                      \displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)

Integration Property [Addition/Subtraction]:                                                          \displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx

Area of a Region Formula:                                                                                     \displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx

Step-by-step explanation:

*Note:

<em>Remember that for the Area of a Region, it is top function minus bottom function.</em>

<u />

<u>Step 1: Define</u>

f(x) = x²

g(x) = x⁶

Bounded (Partitioned) by x-axis

<u>Step 2: Identify Bounds of Integration</u>

<em>Find where the functions intersect (x-values) to determine the bounds of integration.</em>

Simply graph the functions to see where the functions intersect (See Graph Attachment).

Interval: [-1, 1]

Lower bound: -1

Upper Bound: 1

<u>Step 3: Find Area of Region</u>

<em>Integration</em>

  1. Substitute in variables [Area of a Region Formula]:                                     \displaystyle A = \int\limits^1_{-1} {[x^2 - x^6]} \, dx
  2. [Area] Rewrite [Integration Property - Subtraction]:                                     \displaystyle A = \int\limits^1_{-1} {x^2} \, dx - \int\limits^1_{-1} {x^6} \, dx
  3. [Area] Integrate [Integration Rule - Reverse Power Rule]:                           \displaystyle A = \frac{x^3}{3} \bigg| \limit^1_{-1} - \frac{x^7}{7} \bigg| \limit^1_{-1}
  4. [Area] Evaluate [Integration Rule - FTC 1]:                                                    \displaystyle A = \frac{2}{3} - \frac{2}{7}
  5. [Area] Subtract:                                                                                               \displaystyle A = \frac{8}{21}

Topic: AP Calculus AB/BC (Calculus I/II)  

Unit: Area Under the Curve - Area of a Region (Integration)  

Book: College Calculus 10e

6 0
3 years ago
David and Mark are marking exam papers. Each set takes David 24 minutes and Mark 1 hour. Express the times David and Mark take a
attashe74 [19]

Answer:

2:5

Step-by-step explanation:

david=24mins

mark = 1hour

change 1 hour to mins

DAVID=24

MARK=60

24:60

THEN SIMPLIFY

24/60

<em>=</em><em>2</em><em>/</em><em>5</em>

<em><u>2</u></em><em><u>:</u></em><em><u>5</u></em>

8 0
3 years ago
Keisha throws a rock down an old well the distance D in Feet The Rock Falls after T seconds can be represented by D =16t ^ 2 + 6
sesenic [268]

Answer:

<em>1</em><em> </em><em>second</em><em> </em>

Step-by-step explanation:

<em>16t</em><em>^</em><em>2</em><em>+</em><em>64t</em><em>=</em><em>80</em>

<em>16t</em><em>^</em><em>2</em><em>+</em><em>64t-80</em><em>=</em><em>0</em>

<em>64</em><em>^</em><em>2-4</em><em>(</em><em>16</em><em>)</em><em>(</em><em>-80</em><em>)</em>

<em>9216</em>

<em>-64-96</em>

<em>______</em><em> </em><em>=</em><em> </em><em>-5</em>

<em> </em><em> </em><em> </em><em>32</em>

<em>-64</em><em>+</em><em>96</em>

<em>_______</em><em> </em><em>=</em><em> </em><em>1</em><em> </em><em>sec</em>

<em> </em><em> </em><em> </em><em> </em><em>32</em>

7 0
3 years ago
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