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Olegator [25]
3 years ago
6

Aiden refills three token machines in an arcade. He puts twice the number in machine A as in machine B, and in machine C , he pu

ts 3/4 of what he put in machine A. The total amount of tokens was 18324. How many tokens does each machine have.
Mathematics
2 answers:
Sonja [21]3 years ago
7 0
It is a possibility of being 16 or 5/6
LenaWriter [7]3 years ago
4 0

Answer:

Machine A has 8144 tokens, machine B has 4072 tokens and machine C has 6108 tokens.

Step-by-step explanation:

Let call: X the number of tokens that machine A has, Y the number of tokens that machine B has and Z the number of tokens that machine C has.

From the sentence: He puts twice the number in machine A as in machine B, we can formulate the equation 1 as:

X = 2*Y

Solving for Y we can write the equation 1 as:

Y = X/2

From the sentence: in machine C , he puts 3/4 of what he put in machine A, we can formulate the equation 2 as:

Z = (3/4)*X

And from the sentence the total amount of tokens was 18324. we can formulate the equation 3 as:

X + Y + Z = 18324

So, if we replace the Y by X/2 and Z by (3/4)X on equation 3 and solve for X, we get:

X + Y + Z = 18324

X + X/2 + (3/4)X = 18324

(9/4)X=18324

X = 18324*(4/9)

X = 8144

Then, replacing this value on equation 1 and 2 to find Y and Z respectively, we get:

Y = X/2 = 8144/2 = 4071

Z = (3/4)*X = (3/4)*4071 = 6108

So, machine A has 8144 tokens, machine B has 4072 tokens and machine C has 6108 tokens.

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A heavy rope, 50 ft long, weighs 0.6 lb/ft and hangs over the edge of a building 120 ft high. Approximate the required work by a
Anastasy [175]

Answer:

Exercise (a)

The work done in pulling the rope to the top of the building is 750 lb·ft

Exercise (b)

The work done in pulling half the rope to the top of the building is 562.5 lb·ft

Step-by-step explanation:

Exercise (a)

The given parameters of the rope are;

The length of the rope = 50 ft.

The weight of the rope = 0.6 lb/ft.

The height of the building = 120 ft.

We have;

The work done in pulling a piece of the upper portion, ΔW₁ is given as follows;

ΔW₁ = 0.6Δx·x

The work done for the second half, ΔW₂, is given as follows;

ΔW₂ = 0.6Δx·x + 25×0.6 × 25 =  0.6Δx·x + 375

The total work done, W = W₁ + W₂ = 0.6Δx·x + 0.6Δx·x + 375

∴ We have;

W = 2 \times \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= 2 \times \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 750

The work done in pulling the rope to the top of the building, W = 750 lb·ft

Exercise (b)

The work done in pulling half the rope is given by W₂ as follows;

W_2 =  \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 562.5

The work done in pulling half the rope, W₂ = 562.5 lb·ft

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Write 931 x 10-5 in standard notation.
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Answer:

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Step-by-step explanation:

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Answer:

f(g(x)) = 4x² + 16x + 13

Step-by-step explanation:

Given the composition of functions f(g(x)), for which f(x) = 4x + 5, and g(x) = x² + 4x + 2.

<h3><u>Definitions:</u></h3>
  • The <u>polynomial in standard form</u> has terms that are arranged by <em>descending</em> order of degree.
  • In the <u>composition of function</u><em> f  </em>with function <em>g</em><em>, </em>which is alternatively expressed as <em>f  </em>° <em>g,</em> is defined as (<em>f </em> ° <em>g</em>)(x) = f(g(x)).

In evaluating composition of functions, the first step is to evaluate the inner function, g(x). Then, we must use the derived value from g(x) as an input into f(x).

<h3><u>Solution:</u></h3>

Since we are not provided with any input values to evaluate the given composition of functions, we can express the given functions as follows:

f(x) = 4x + 5

g(x) = x² + 4x + 2

f(g(x)) = 4(x² + 4x + 2)  + 5

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f(g(x)) = 4x² + 16x + 8  + 5

Combine constants:

f(g(x)) = 4x² + 16x + 13

Therefore, f(g(x)) as a polynomial in <em>x</em> that is written in standard form is: 4x² + 16x + 13.

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