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notsponge [240]
2 years ago
13

A store is selling a black pair of shoes for $79.45 and a brown pair of shoes for $133.99. the black pair of shoes is marked up

by 44%, and the brown pair of shoes is marked down by 21%. which pair of shoes is more expensive, and how much more expensive is it? round all dollar values to the nearest cent.
Mathematics
2 answers:
borishaifa [10]2 years ago
6 0

Answer:

Its B on edge

Step-by-step explanation:

Aleks [24]2 years ago
5 0

let:
price after markup black shoes = x
price after markdown brown shoes=y
where
a black  pair of shoe price= $79.45
a brown pair of shoe price =133.99
solve for x and y
x= $79.45 x(1+0.44)
x=114.41
y=$133.99(1-0.21)
y=$105.85
The black pair of shoes is more expensive than the brown pair of shoes.
difference= x-y
difference=$114.41-$105.85
difference =$8.56
The black pair of shoes is expensive by $8.56.  


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7 0
2 years ago
She bought <br> 314yards of fabric. Her total cost was $13. What was the cost per yard?
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Step-by-step explanation:

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2 years ago
Maria's height while jumping on a a trampoline can be modeled by the equation h=-16t^2+18t+5. Where t=time in seconds and h=heig
Vadim26 [7]

Answer:

10.06 ft

Step-by-step explanation:

Maria's maximum height will occur when her velocity reaches zero (0). This means that she has stopped ascending and is about to begin descent.

The equation for the height reached by Maria on the trampoline is given as:

h=-16t^2+18t+5

To find her maximum height, we first have to find the time it will take her to get to that height and corresponding velocity (zero).

Her velocity can be found by differentiating her height i.e. dh/dt:

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Therefore, when v = 0:

0 = -32t+ 18\\\\=> 32t= 18\\\\t = 18 / 32 = 0.5625 secs

It takes her 0.5625 seconds to get to her maximum height.

Therefore, her height at that time (0.5625 seconds) is:

h=-16(0.5625)^2+18(0.5625)+5\\\\h = -16 * (0.3164) + 10.125+5\\\\h = -5.0624 + 15.125\\\\h = 10.06 ft

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3 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
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