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Alex Ar [27]
3 years ago
13

A: 2x + 8y = 5 B: x + 9y = 10 To solve this system of equations by elimination, what could you multiply equation B by if you wan

t to add the equations together and have a resulting equation with only the y-variable (the x-variable is eliminated)?
Mathematics
2 answers:
Leona [35]3 years ago
6 0

Answer:

-2

Step-by-step explanation:

shepuryov [24]3 years ago
3 0
Multiply equation B by 2, so then it is 2x +18y = 20, then you can eliminate x and leaving to solve for y
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What is the derivative of 1/square root 4x.
Bumek [7]

Answer:

\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4x^\bigg{\frac{3}{2}}}

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>

Exponential Properties

  • 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 Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

\displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg]

<u>Step 2: Differentiate</u>

  1. Simplify:                                                                                                         \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \bigg( \frac{1}{2\sqrt{x}} \bigg)'
  2. Rewrite [Derivative Property - Multiplied Constant]:                                   \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{1}{\sqrt{x}} \bigg)'
  3. Rewrite [Exponential Rule - Root Rewrite]:                                                 \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{1}{x^\Big{\frac{1}{2}}} \bigg)'
  4. Rewrite [Exponential Rule - Rewrite]:                                                           \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( x^\bigg{\frac{-1}{2}} \bigg)'
  5. Derivative Rule [Basic Power Rule]:                                                             \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{1}{2} \bigg( \frac{-1}{2} x^\bigg{\frac{-3}{2}} \bigg)
  6. Simplify:                                                                                                         \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4} x^\bigg{\frac{-3}{2}}
  7. Rewrite [Exponential Rule - Rewrite]:                                                           \displaystyle \frac{d}{dx} \bigg[ \frac{1}{\sqrt{4x}} \bigg] = \frac{-1}{4x^\bigg{\frac{3}{2}}}

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

Unit: Differentiation

5 0
3 years ago
How to estimate the quotient of 57.8÷81
monitta
Round 57.8 to 60 and round 81 to 80.

60/80 or, if you simplify it, 3/4

And it is common knowledge that 3/4 = .75

So the quotient of 57.8/81 is ~.75


3 0
3 years ago
Read 2 more answers
Help please pretty please i do not understand
Natalija [7]
If i remember correctly i think you just cross multiply   <span />
7 0
3 years ago
Read 2 more answers
the constant value which is repeatedly added to each term in an arithmetic sequence to obtain the next term​
Mariulka [41]

Answer:

<h2>This value is called the common difference</h2>

Step-by-step explanation:

The common difference is the constant value which is repeatedly added to each term in an arithmetic sequence to obtain the next term​, it is basically the difference between consecutive numbers

To find the common difference we can subtract the previous term from the first time or the second to the last term from the last term, the idea of finding the common difference is basically subtracting the previous term form the subsequent term.

7 0
4 years ago
What is the simplified form of -5 to the -2 power?
SOVA2 [1]
The answer is 10 hope this helps
4 0
3 years ago
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