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3 years ago

If possible, prove that AB||CD. If not possible, show and explain why

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

8 0

Answer:

AB || CD

Step-by-step explanation:

Remember that vertical angles are always equal/ congruent to each other

Since 108° is vertical to angle S, angle S also equals 108°

Since a line is equal to 180°, and you now know that S equals 108°, you would subtract 180 by 108 to get line RT.

180-108= 71°

By default, line VW also equals 71° (you are repeating the same progress you did to get RT)

The angle vertical to <V is equal to 134°, so angle V also equals 134°

180-134 gives you the answer to the angle by 134 and V, they should equal 46°

Corresponding angles equal each other, so angle R is also 134°, and by default the other numbers are equal to to what you did at the top right

Once again, since corresponding angles are equal to each other, you would automatically have the same answers for the bottom line;

the outermost angles equal 134° and the innermost angles equal 46°

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Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

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3 years ago

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1. Write an equation in slope-intercept form for a line passing through (-3,3) with a slope of 1.

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$\bf (\stackrel{x_1}{-4}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{12}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{12-2}{1-(-4)}\implies \cfrac{10}{1+4}\implies \cfrac{10}{5}\implies 2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=2[x-(-4)]\implies y-2=2(x+4) \\\\\\ y-2=2x+8\implies y=2x+10$

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4 years ago

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Answer:

Option A.

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Graph of function has been given in the figure.

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x -2 -1 0 1 2

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Therefore, coordinates of the points which will lie on the inverse of this function will be

x -10 -3 -2 -1 6

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Therefore, Option A. will be the answer.

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3 years ago

I will mark Brainliest!! 87 Points!!

DIA [1.3K]

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shift to the left 1 unit (replace x with x+1)
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3 years ago

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Express 3/10 + 45/100 as a fraction with denominator 100

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3/10 + 45/100
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If possible, prove that AB||CD. If not possible, show and explain why​

If possible, prove that AB||CD. If not possible, show and explain why