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77julia77 [94]
4 years ago
10

When Scott woke up this morning, the temperature was 2°F. The temperature fell 5° by noon. Scott used this number line to model

the temperature drop.
What error did Scott make?

A. Scott should have started by moving from 0 to –5.

B. Scott should have started by moving from 0 to –2.

C. Scott started correctly by moving from 0 to 2, but then he should have moved 5 units to the left.

D. Scott started correctly by moving from 0 to 2, but then he should have moved 5 units to the right.

Mathematics
2 answers:
Mariulka [41]4 years ago
8 0

Answer:

c

Step-by-step explanation:

I am Lyosha [343]4 years ago
7 0
The answer is C because "dropped" means minus so it is asking: 2-5 which should be -3 not -5.
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Digiron [165]

Answer: (\dfrac13,-2),\ (0,-3),\ (-2,1),\ (-1,1) .

Step-by-step explanation:

Transformation rule for dilation:

(x,y)\to(kx,ky) , where k = scale factor

Given : Scale factor = -\dfrac13

Parallelogram JKLM has vertices J(-1, 6), K(0, 9), L(6, −3), and M(3, −3)

Vertices after dilation:

(-1,6)\to(-1\times\dfrac{-1}{3},6\times\dfrac{-1}{3})=(\dfrac13,-2)

(0,9)\times(0\times-\dfrac13,9\times-\dfrac13)=(0,-3)

(6,-3)\to(6\times-\dfrac13,\ -3\times-\dfrac13)=(-2,1)

(3,-3)\to (3\times-\dfrac13,\ -3\times-\dfrac13)=(-1,1)

Hence, the coordinates of the image if the parallelogram = (\dfrac13,-2),\ (0,-3),\ (-2,1),\ (-1,1) .

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Anvisha [2.4K]

Answer:

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General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
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The derivative of a constant is equal to 0

Basic Power Rule:

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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)

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<u>Step 1: Define</u>

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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}
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