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kiruha [24]
3 years ago
8

(Quick please ) !!!!!!!!!!!!!!!! Help me!!!!! find the surface area of the rectangular prism!!!!!! thank you

Mathematics
1 answer:
Aleks [24]3 years ago
3 0
You can do this !

The rectangular prism has

                   -- length = 10 cm
                   -- width  =  7 cm
                   -- height =  3 cm.

-- The area of the top and bottom is  (length x width)  each.

-- The area of the left and right sides is  (length x height)  each.

-- The area of the front and back is  (width x height)  each.

There.  I just laid out all the schmartz you need to answer this question.
The rest is all simple arithmetic, and you're perfectly capable of turning
the crank and getting the answer.  You don't need anybody else to do
that part for you.

Don't forget your units.  The area of each flat face is (cm) times (cm),
and that product will be some cm² , for area.
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If you were to use the substitution method to solve the following system, choose the new equation after the expression equivalen
Alisiya [41]
The answer is <span>2(–4y + 13) – 3y = –29

Step 1: Express </span><span>x from the second equation
Step 2: Substitute x into the first equation:

The system of equations is:
</span><span>2x – 3y = –29 
x + 4y = 13

Step 1: 
</span>The second equation is:    x + 4y = 13
Rearrange it to get x:         x = - 4y + 13

Step 2:
The first equation is:          2x – 3y = –29 
The second equation is:    x = - 4y + 13
Substitute x from the second equation into the first one:
2(-4y + 13) - 3y = -29

Therefore, the second choice is correct.
5 0
3 years ago
The curve
kherson [118]

Answer:

Point N(4, 1)

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  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Algebra I</u>

  • Coordinates (x, y)
  • Functions
  • Function Notation
  • Terms/Coefficients
  • Anything to the 0th power is 1
  • Exponential Rule [Rewrite]:                                                                              \displaystyle b^{-m} = \frac{1}{b^m}
  • Exponential Rule [Root Rewrite]:                                                                     \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

  • f(x) = cxⁿ
  • 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:

<u>Step 1: Define</u>

<u />\displaystyle y = \sqrt{x - 3}<u />

<u />\displaystyle y' = \frac{1}{2}<u />

<u />

<u>Step 2: Differentiate</u>

  1. [Function] Rewrite [Exponential Rule - Root Rewrite]:                                   \displaystyle y = (x - 3)^{\frac{1}{2}}
  2. Chain Rule:                                                                                                        \displaystyle y' = \frac{d}{dx}[(x - 3)^{\frac{1}{2}}] \cdot \frac{d}{dx}[x - 3]
  3. Basic Power Rule:                                                                                             \displaystyle y' = \frac{1}{2}(x - 3)^{\frac{1}{2} - 1} \cdot (1 \cdot x^{1 - 1} - 0)
  4. Simplify:                                                                                                             \displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}} \cdot 1
  5. Multiply:                                                                                                             \displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}}
  6. [Derivative] Rewrite [Exponential Rule - Rewrite]:                                          \displaystyle y' = \frac{1}{2(x - 3)^{\frac{1}{2}}}
  7. [Derivative] Rewrite [Exponential Rule - Root Rewrite]:                                 \displaystyle y' = \frac{1}{2\sqrt{x - 3}}

<u>Step 3: Solve</u>

<em>Find coordinates</em>

<em />

<em>x-coordinate</em>

  1. Substitute in <em>y'</em> [Derivative]:                                                                             \displaystyle \frac{1}{2} = \frac{1}{2\sqrt{x - 3}}
  2. [Multiplication Property of Equality] Multiply 2 on both sides:                      \displaystyle 1 = \frac{1}{\sqrt{x - 3}}
  3. [Multiplication Property of Equality] Multiply √(x - 3) on both sides:            \displaystyle \sqrt{x - 3} = 1
  4. [Equality Property] Square both sides:                                                           \displaystyle x - 3 = 1
  5. [Addition Property of Equality] Add 3 on both sides:                                    \displaystyle x = 4

<em>y-coordinate</em>

  1. Substitute in <em>x</em> [Function]:                                                                                \displaystyle y = \sqrt{4 - 3}
  2. [√Radical] Subtract:                                                                                          \displaystyle y = \sqrt{1}
  3. [√Radical] Evaluate:                                                                                         \displaystyle y = 1

∴ Coordinates of Point N is (4, 1).

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

Unit: Derivatives

Book: College Calculus 10e

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Step-by-step explanation:

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Answer: A B and C

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