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Brrunno [24]
3 years ago
7

Two customers went to a post office to buy postcards and large envelopes. Each postcard

Mathematics
1 answer:
astraxan [27]3 years ago
4 0

The right answer is Option A.

Step-by-step explanation:

Let,

x be the postcards

y be the large envelops

According to given statement;

14x+5y=12   Eqn 1

10x+15y=24.80   Eqn 2

Multiplying Eqn 1 by 3;

3(14x+5y=12)\\42x+15y=36\ \ \ Eqn \ 3

Subtracting Eqn 2 from Eqn 3;

(42x+15y)-(10x+15y)=36-24.80\\42x+15y-10x-15y=11.20\\32x=11.20\\

Dividing both sides by 32

\frac{32x}{32}=\frac{11.20}{32}\\x=0.35

Putting x=0.35 in Eqn 1;

14(0.35)+5y=12\\4.90+5y=12\\5y=12-4.90\\5y=7.10

Dividing both sides by 5

\frac{5y}{5}=\frac{7.10}{5}\\y=1.42

Therefore, one large envelope costs $1.42

The right answer is Option A.

Keywords: linear equations, subtraction

Learn more about linear equations at:

  • brainly.com/question/9738996
  • brainly.com/question/10015690

#LearnwithBrainly

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Anvisha [2.4K]

Answer:

y''(-1) =8

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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  2. Parenthesis
  3. Exponents
  4. Multiplication
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  • Left to Right

Equality Properties

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Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

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  • 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>

-xy - 2y = -4

Rate of change of the tangent line at point (-1, 4)

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

<em>Find 1st Derivative</em>

  1. Implicit Differentiation [Product Rule/Basic Power Rule]:                            -y - xy' - 2y' = 0
  2. [Algebra] Isolate <em>y'</em> terms:                                                                               -xy' - 2y' = y
  3. [Algebra] Factor <em>y'</em>:                                                                                       y'(-x - 2) = y
  4. [Algebra] Isolate <em>y'</em>:                                                                                         y' = \frac{y}{-x-2}
  5. [Algebra] Rewrite:                                                                                           y' = \frac{-y}{x+2}

<u>Step 3: Find </u><em><u>y</u></em>

  1. Define equation:                    -xy - 2y = -4
  2. Factor <em>y</em>:                                 y(-x - 2) = -4
  3. Isolate <em>y</em>:                                 y = \frac{-4}{-x-2}
  4. Simplify:                                 y = \frac{4}{x+2}

<u>Step 4: Rewrite 1st Derivative</u>

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  2. [Algebra] Simplify:                                                                                         y' = \frac{-4}{(x+2)^2}

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<em>Find 2nd Derivative</em>

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Answer:

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Step-by-step explanation:

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