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yawa3891 [41]
2 years ago
6

The volume of a rectangular box can be found using the formula lwh where l represents the length, w represents the width, and h

represents the height of the box. What is the volume of a box with the following dimensions?
I = 4 centimeters
W= 5 centimeters
H=6 centimeters
Mathematics
1 answer:
Dafna1 [17]2 years ago
6 0

Answer:

120 cm^3

Step-by-step explanation:

volume = lwh

I = 4 centimeters  

w = 5 centimeters  

h = 6 centimeters

volume = (4 cm)(5 cm)(6 cm)

volume = 120 cm^3

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Answer:

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A certain store sells small, medium and large toy trucks in each of the colors red blue green and yellow. the store has an equal
tangare [24]
There is a 1/2 chance that he will get at least one of the two features.

There are 3 x 4 = 12 possibilities. There are all 4 colors of each 3 sizes.

All 4 of the medium trucks have one of the features.

For the other 2 sizes, only the red truck will have one of the features.

There are 4 + 2 = 6 possibilities out of 12 for a chance of 6/12 or 1/2.
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3 years ago
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PLEASE HELP ME WITH THIS QUESTION PLEASE
Ugo [173]

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a. 2.2 m

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6 0
2 years ago
Please help ASAP!!!??
Mazyrski [523]

Answer:

\{-2,-1,1,2,4\}

Step-by-step explanation:

Given

The attached graph

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First, we list out the coordinate of each point on the graph:

The points are:

\{(-4,-1),(-3,-2),(-2,2),(1,1),(2,4)\}

A function has the form: (x,y)

Where

y = range:

From the coordinate points above,

y = \{-1,-2,2,1,4\}

Order from least to greatest"

y = \{-2,-1,1,2,4\}

Hence, the range are: \{-2,-1,1,2,4\}

4 0
2 years ago
I guess I'm lacking in differential equations. I couldn't solve this question. Can you help me?
Sonja [21]

Answer:

See Explanation.

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Equality Properties
  • Reciprocals

<u>Algebra II</u>

  • Log/Ln Property: ln(\frac{a}{b} ) = ln(a) - ln(b)

<u>Calculus</u>

Derivatives

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

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

Derivative of Ln: \frac{d}{dx} [ln(u)] = \frac{u'}{u}

Step-by-step explanation:

<u>Step 1: Define</u>

ln(\frac{2x-1}{x-1} )=t

<u>Step 2: Differentiate</u>

  1. Rewrite:                                                                                                         t = ln(\frac{2x-1}{x-1})
  2. Rewrite [Ln Properties]:                                                                                 t = ln(2x-1) - ln(x - 1)
  3. Differentiate [Ln/Chain Rule/Basic Power Rule]:                                         \frac{dt}{dx} = \frac{1}{2x-1} \cdot 2 - \frac{1}{x-1} \cdot 1
  4. Simplify:                                                                                                          \frac{dt}{dx} = \frac{2}{2x-1} - \frac{1}{x-1}
  5. Rewrite:                                                                                                          \frac{dt}{dx} = \frac{2(x-1)}{(2x-1)(x-1)} - \frac{2x-1}{(2x-1)(x-1)}
  6. Combine:                                                                                                       \frac{dt}{dx} = \frac{-1}{(2x-1)(x-1)}
  7. Reciprocate:                                                                                                  \frac{dx}{dt} = -(2x-1)(x-1)
  8. Distribute:                                                                                                         \frac{dx}{dt} = (1-2x)(x-1)
8 0
2 years ago
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