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taurus [48]
3 years ago
5

The scoop of an excavator holds 27 ft3 of sand. If the capacity of a freight car is 40 tons of sand and we know that 1 ft3 of sa

nd weighs 0.5 tons, how many freight cars can be filled with 120 loads?
Mathematics
2 answers:
kozerog [31]3 years ago
5 0
27ft³=x tons
1ft³=0.5 tons

\frac{27cubicfeet}{x_t_o_n_s} = \frac{1cubic foot}{0.5 tons}

Cross multiply
1x=13.5tons
x=13.5tons

So one scoop/load carries 13.5 tons of sand. Lets see how many 120 loads can carry.

13.5tons*120loads=1620 tons of sand

Now lets see how freight cars it would take to hold 1620 tons of sand

1620/40=40.5 freight cars

Since you can't have .5 of a freight car, lets round 40.5 up to 41

Answer=41 freight cars

Rus_ich [418]3 years ago
3 0

Answer:

40.5

Step-by-step explanation:

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Circle O has a circumference of approximately 44 in.
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Step-by-step explanation:

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3 years ago
Rewrite the expression as a multiple of a sum of two numbers with no common factor. <br> 30 + 12
fomenos
The expression 30 + 12 can be rewritten as 2x(13+8)
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4 years ago
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What is the average rate of change of the function g(x) = 4x from x = 1 to x = 5?
Fittoniya [83]

Answer:

\displaystyle 4

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>

  • Function Notation
  • Interval Notation - [a, b]
  • Average Rate of Change: \displaystyle \displaystyle \frac{f(b) - f(a)}{b - a}

Step-by-step explanation:

<u>Step 1: Define</u>

g(x) = 4x

Interval [1, 5]

<u>Step 2: Find Change</u>

  1. Substitute in function and points [Average Rate of Change]:                     \displaystyle \displaystyle \frac{4(5) - 4(1)}{5 - 1}
  2. [Fraction] Multiply:                                                                                          \displaystyle \displaystyle \frac{20 - 4}{5 - 1}
  3. [Fraction] Subtract:                                                                                         \displaystyle \displaystyle \frac{16}{4}
  4. [Fraction] Divide:                                                                                              \displaystyle 4
7 0
3 years ago
Let a and ß be first quadrant angles with cos(a)=
Finger [1]

Since both α and β are in the first quadrant, we know each of cos(α), sin(α), cos(β), and sin(β) are positive. So when we invoke the Pythagorean identity,

sin²(x) + cos²(x) = 1

we always take the positive square root when solving for either sin(x) or cos(x).

Given that cos(α) = √11/7 and sin(β) = √11/4, we find

sin(α) = √(1 - cos²(α)) = √38/7

cos(β) = √(1 - sin²(β)) = √5/4

Now, recall the sum identity for cosine,

cos(x + y) = cos(x) cos(y) - sin(x) sin(y)

It follows that

cos(α + β) = √11/7 × √5/4 - √38/7 × √11/4 = (√55 - √418)/28

7 0
2 years ago
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