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dezoksy [38]
2 years ago
8

The Cooper family, consisting of 6 children and 2 adults, goes to the movies and pays a total of $86. The Griffen birthday party

, consisting of 10 children and 3 adults, pays a total of $139 for tickets. Let x = the cost of a child ticket and y = the cost of an adult ticket. How much does a child's ticket cost and how much does an adult ticket cost?
Mathematics
1 answer:
Oksanka [162]2 years ago
4 0

Answer: The cost of child's ticket = $10

The cost of adult ticket = $ 13

Step-by-step explanation:

Let x be the cost of a child ticket and y be the cost of an adult ticket.

Then According to the question, we have

6x+2y=86..........................(1)\\\\10x+3y=139.......................(2)

Multiply equation (1) by 3 and equation (2) by 2, then we have

18x+6y=258.......................(1)\\\\20x+6y=278...........................(2)

Subtract equation (1) from equation (2), we have

2x=20\\\\\Rightarrow\ x=10

Substitute the value of x in equation (1), we get

60+2y=86\\\\\Rightarrow\ 2y=26\\\\\Rightarrow\ y=13

Hence, the cost of child's ticket = $10

The cost of adult ticket = $ 13

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Blake is a member of a golf club. When he became a member, he deposited money into his club account. For each round of golf that
Leno4ka [110]

Answer:

D)  Blake's account initially had $320, which is decreasing at the rate of $20 per round.

Step-by-step explanation:

Here, according to the question:

x: The total number of golf rounds played

y: Represents the amount, in dollars, left in his account after playing x rounds

Let as assume A is the amount deposited initially.

y = A  - n x    here n : rate of each round played

Also, here

if x  = 2, then y = 280

if x  =4, then y = 240

If x  = 2 and y = 280, then:

y = A  - n x ⇒  280 = A  - 2 n   ... (1)

If x  = 4 and y = 240, then:

y = A  - n x ⇒  240 = A  - 4 n   ... (2)

Solving (1) and (2)  by (1) - (2),  we get:

40 = 2 n or, n  = 20

Putting n = 20 in 280 = A  - 2 n, we get: A  = 280 +  2(20) = 320

⇒ A = $320, n = $20

⇒ The amount initially deposited is $320 and the rate per round is  $20

Hence ,Blake's account initially had $320, which is decreasing at the rate of $20 per round.

4 0
2 years ago
Least common multiple of 9(x+2)(2x-1) and 3x+2
mixas84 [53]

Answer:

LCM(9, -1, -1, +2, +2, +2) = 2×3^2 = 18

Step-by-step explanation:

9 = 32

-1 = -1

-1 = -1

+2 = 2

+2 = 2

+2 = 2

LCM = 2×3^2 = 18

18 / 9 = 2

18 / -1 = -18

18 / -1 = -18

18 / +2 = 9

18 / +2 = 9

18 / +2 = 9

3 0
3 years ago
Hello again! This is another Calculus question to be explained.
podryga [215]

Answer:

See explanation.

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>

Functions

  • Function Notation
  • Exponential Property [Rewrite]:                                                                   \displaystyle b^{-m} = \frac{1}{b^m}
  • Exponential Property [Root Rewrite]:                                                           \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           \displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)

Derivative Property [Addition/Subtraction]:                                                         \displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

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

Step-by-step explanation:

We are given the following and are trying to find the second derivative at <em>x</em> = 2:

\displaystyle f(2) = 2

\displaystyle \frac{dy}{dx} = 6\sqrt{x^2 + 3y^2}

We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

\displaystyle \frac{dy}{dx} = 6(x^2 + 3y^2)^\big{\frac{1}{2}}

When we differentiate this, we must follow the Chain Rule:                             \displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]

Use the Basic Power Rule:

\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} (2x + 6yy')

We know that y' is the notation for the 1st derivative. Substitute in the 1st derivative equation:

\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big]

Simplifying it, we have:

\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]

We can rewrite the 2nd derivative using exponential rules:

\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}

To evaluate the 2nd derivative at <em>x</em> = 2, simply substitute in <em>x</em> = 2 and the value f(2) = 2 into it:

\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}

When we evaluate this using order of operations, we should obtain our answer:

\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = 219

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

Unit: Differentiation

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2 years ago
At the beginning of the week, Paloma opened a new box of 250 coffee creamers. Forty-four coffee creamers were used during the we
LiRa [457]
Hello, the correct answer would be B. Because if she had 250 then used 44 you would subtract 44 from 250 to find x

Hope this helps :)
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You can find them by substituting y for 0
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