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omeli [17]
3 years ago
5

John wants to fence in a portion of his backyard. ... How many feet of fencing does john need to buy to enclose this area? ... H

ow many square yards of carpet will she need to complete the room?
Mathematics
1 answer:
Lelechka [254]3 years ago
3 0
For this problem you need to find the area.
To find area, you multiply the width by the length.
You multiply 12 1/2 by 5 1/2 and you get 31 1/4
This is how you do it::

= 25/2 × 5/2

= (25 × 5) / (2 × 2)

= 125/4

= 125/4

<span>= 31 1/4

So you know that the area is 31 1/4 yards. 
In the question it asks HOW MANY FEET so we have to convert 31 1/4 yards to feet. 
1 yard= 3 feet
31 1/4 *3= </span><span>93.75
<span>John need to buy 93.75 feet of fencing to enclose the area.</span></span>
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If -y-2x^3=Y^2 then find D^2y/dx^2 at the point (-1,-2) in simplest form
algol13

Answer:

\frac{d^2y}{dx^2} = \frac{-4}{3}

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>Algebra I</u>

  • Factoring

<u>Calculus</u>

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Product Rule: \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)

Chain Rule: \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Quotient Rule: \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}

Step-by-step explanation:

<u>Step 1: Define</u>

-y - 2x³ = y²

Rate of change of tangent line at point (-1, -2)

<u>Step 2: Differentiate Pt. 1</u>

<em>Find 1st Derivative</em>

  1. Implicit Differentiation [Basic Power Rule]:                                                  -y'-6x^2=2yy'
  2. [Algebra] Isolate <em>y'</em> terms:                                                                              -6x^2=2yy'+y'
  3. [Algebra] Factor <em>y'</em>:                                                                                       -6x^2=y'(2y+1)
  4. [Algebra] Isolate <em>y'</em>:                                                                                         \frac{-6x^2}{(2y+1)}=y'
  5. [Algebra] Rewrite:                                                                                           y' = \frac{-6x^2}{(2y+1)}

<u>Step 3: Differentiate Pt. 2</u>

<em>Find 2nd Derivative</em>

  1. Differentiate [Quotient Rule/Basic Power Rule]:                                          y'' = \frac{-12x(2y+1)+6x^2(2y')}{(2y+1)^2}
  2. [Derivative] Simplify:                                                                                       y'' = \frac{-24xy-12x+12x^2y'}{(2y+1)^2}
  3. [Derivative] Back-Substitute <em>y'</em>:                                                                     y'' = \frac{-24xy-12x+12x^2(\frac{-6x^2}{2y+1} )}{(2y+1)^2}
  4. [Derivative] Simplify:                                                                                      y'' = \frac{-24xy-12x-\frac{72x^4}{2y+1} }{(2y+1)^2}

<u>Step 4: Find Slope at Given Point</u>

  1. [Algebra] Substitute in <em>x</em> and <em>y</em>:                                                                     y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(-1)^4}{2(-2)+1} }{(2(-2)+1)^2}
  2. [Pre-Algebra] Exponents:                                                                                      y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(1)}{2(-2)+1} }{(2(-2)+1)^2}
  3. [Pre-Algebra] Multiply:                                                                                   y''(-1,-2) = \frac{-48+12-\frac{72}{-4+1} }{(-4+1)^2}
  4. [Pre-Algebra] Add:                                                                                         y''(-1,-2) = \frac{-36-\frac{72}{-3} }{(-3)^2}
  5. [Pre-Algebra] Exponents:                                                                               y''(-1,-2) = \frac{-36-\frac{72}{-3} }{9}
  6. [Pre-Algebra] Divide:                                                                                      y''(-1,-2) = \frac{-36+24 }{9}
  7. [Pre-Algebra] Add:                                                                                          y''(-1,-2) = \frac{-12}{9}
  8. [Pre-Algebra] Simplify:                                                                                    y''(-1,-2) = \frac{-4}{3}
6 0
3 years ago
Identify the correct ways to classify ∠1 and ∠4 in the figure shown below.
lapo4ka [179]
They are adjacent and complementary.
Next to each other and add up to 90 degrees.
4 0
3 years ago
Read 2 more answers
Carmen Martinez
klemol [59]
  • (4,4)
  • (10,7)

\boxed{\sf Slope(m)=\dfrac{y_2-y_1}{x_2-x_1}}

\\ \sf\longmapsto m=\dfrac{7-4}{10-4}

\\ \sf\longmapsto m=\dfrac{3}{6}

\\ \sf\longmapsto m=\dfrac{1}{2}

\\ \sf\longmapsto m\approx0.5

6 0
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Solve. d + 8 &lt; 35 <br> d &lt; –27 <br> d &lt; 27 <br> d &lt; –43 <br> d &lt; 43
Sindrei [870]
The answer is d lol have a good day
6 0
3 years ago
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Sorry if its blurry im on my laptop
Kamila [148]

Answer:

Step-by-step explanation:

1 in=2 ft

Bedroom 1

scale

17.5 in ×12.5in

actual

(17.5×2)×(12.5×2)

or

35 ft×25 ft

Living Room

scale

17.5 in×17.5 in

actual

35 ft×35 ft

Bathroom 1

scale

12.5 in×12.5 in

actual

25 ft×25 ft

Kitchen

scale

17.5 in ×12.5 in

actual

35 ft×35 ft

Bedroom2

scale

12.5 in ×12.5 in

actual

25 ft×25 ft

Entryway

scale

12.5 in×12.5 in

actual

25 ft×25 ft

Bathroom 2

scale

12.5 in×5 in

actual

25 ft×10 ft

6 0
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