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goldfiish [28.3K]
2 years ago
5

PLEASE ANSWER THIS? WHAT IS EG

Mathematics
1 answer:
crimeas [40]2 years ago
5 0
It's the points of the triangle E----------G
They are the twin points
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− 6 + 3 0 − 4

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1/4+2/3-1/2= how do I solve this
Ann [662]

Answer:

5/12

Step-by-step explanation:

First you add 1/4+2/3.

You need to change the denominator so that they are equal to each other

If you multiply the 1/4 by 3/3 you will get 3/12 which is equal to 1/4

Then you need to do the same to 2/3. This time you need to multiply it by 4/4 to get the same denominator which will be 2/3*4/4=8/12

Then you add 3/12+8/12=11/12. You don't need to add the denominator.

After this you will need to subtract 11/12 and 1/2. This time you need to only change the denominator for 1/2.

You multiply the denominator and numerator by 6/6 to 1/2 and you will get 6/12 then you are going to subtract 11/12-6/12 and you will get 5/12

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3 years ago
Help me BRAINLIEST i dont know
natka813 [3]

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38 and 57

Step-by-step explanation:

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3 years ago
What matrix results from −8·A?
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Implicit differentiation Please help
Anvisha [2.4K]

Answer:

y''(-1) =8

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

<u>Algebra I</u>

  • Factoring

<u>Calculus</u>

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

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

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

Quotient Rule: \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}

Step-by-step explanation:

<u>Step 1: Define</u>

-xy - 2y = -4

Rate of change of the tangent line at point (-1, 4)

<u>Step 2: Differentiate Pt. 1</u>

<em>Find 1st Derivative</em>

  1. Implicit Differentiation [Product Rule/Basic Power Rule]:                            -y - xy' - 2y' = 0
  2. [Algebra] Isolate <em>y'</em> terms:                                                                               -xy' - 2y' = y
  3. [Algebra] Factor <em>y'</em>:                                                                                       y'(-x - 2) = y
  4. [Algebra] Isolate <em>y'</em>:                                                                                         y' = \frac{y}{-x-2}
  5. [Algebra] Rewrite:                                                                                           y' = \frac{-y}{x+2}

<u>Step 3: Find </u><em><u>y</u></em>

  1. Define equation:                    -xy - 2y = -4
  2. Factor <em>y</em>:                                 y(-x - 2) = -4
  3. Isolate <em>y</em>:                                 y = \frac{-4}{-x-2}
  4. Simplify:                                 y = \frac{4}{x+2}

<u>Step 4: Rewrite 1st Derivative</u>

  1. [Algebra] Substitute in <em>y</em>:                                                                               y' = \frac{-\frac{4}{x+2} }{x+2}
  2. [Algebra] Simplify:                                                                                         y' = \frac{-4}{(x+2)^2}

<u>Step 5: Differentiate Pt. 2</u>

<em>Find 2nd Derivative</em>

  1. Differentiate [Quotient Rule/Basic Power Rule]:                                          y'' = \frac{0(x+2)^2 - 8 \cdot 2(x + 2) \cdot 1}{[(x + 2)^2]^2}
  2. [Derivative] Simplify:                                                                                      y'' = \frac{8}{(x+2)^3}

<u>Step 6: Find Slope at Given Point</u>

  1. [Algebra] Substitute in <em>x</em>:                                                                               y''(-1) = \frac{8}{(-1+2)^3}
  2. [Algebra] Evaluate:                                                                                       y''(-1) =8
6 0
3 years ago
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