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Nat2105 [25]
4 years ago
13

Which of the following steps were applied to ABCD to obtain A'B'C'D'?

Mathematics
1 answer:
alisha [4.7K]4 years ago
5 0

It was reflected, when a figure is reflected, all points along each segment are reflected. In other words, a point lying on AB in the pre-image will lie on A'B' in the image. If point E is the midpoint of AB, then point E' will be the midpoint of A'B'. Therefore, a reflected figure preserves collinearity and betweenness of points, just like with translated figures. Also, if a line of symmetry exists in the pre-image, it will exist in the image.

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ad-work [718]
Supplementary means the 2 angles add up to 180 so...

124+(2x+4)=180

124+2x+4=180

124+2x+4-4=180-4

124+2x=176

124-124+2x=176-124

2x=52

2x/2=52/2

x=26
plug in and check

124+2x+4=180

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180=180


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3 years ago
Solving Exponential and Logarithmic Equations In Exercise, solve for x.<br> In 2x - In(3x - 1) = 0
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Answer:

\frac{-1}{3x^2-x}

Step-by-step explanation:

  1. If f(x) is in th form of f(x)=g(x)-h(x) then f'(x)=g'(x) - h'(x)
  2. When f(x)=z(g(x)) then f'(x)= z'(g(x))g'(x) (called as chain rule)

<u>using these information</u>:

g(x)=ln2x then g'(x)=\frac{(2x)'}{2x} =\frac{2}{2x}=\frac{1}{x}

h(x)=In(3x - 1) then h'(x)=\frac{(3x-1)'}{3x-1} =\frac{3}{3x-1}f'(x)=g'(x) - h'(x) =[tex]\frac{1}{x} - \frac{3}{3x-1} =\frac{-1}{3x^2-x}

7 0
4 years ago
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