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Naddik [55]
3 years ago
11

Can anyone help me with this problem ots substitution in math

Mathematics
1 answer:
andrew11 [14]3 years ago
3 0
X - 3y = 12.....x = 3y + 12
so sub in 3y + 12 in for x in the other equation

2x + 5y = -20
2(3y + 12) + 5y = -20
6y + 24 + 5y = -20
6y + 5y = -20 - 24
11y = - 44
y = -44/11
y = -4

x - 3y = 12
x - 3(-4) = 12
x + 12 = 12
x = 12 - 12
x = 0

solution is (0,-4)
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Answer:

(a)  See below.

(b)  x = 0 or x = 1

(c)  x = 0 removable, x = 1 non-removable

Step-by-step explanation:

Given rational function:

f(x)=\dfrac{\ln |x-1|}{x}

<u>Part (a)</u>

Substitute x = 2 into the given rational function:

\begin{aligned}\implies f(2) & =\dfrac{\ln |2-1|}{2}\\\\ & =\dfrac{\ln 1}{2}\\\\ & =\dfrac{0}{2}\\\\ & =0\end{aligned}

Therefore, as the function is defined at x = 2, the function is continuous at x = 2.

<u>Part (b)</u>

Given interval:  [-2, 2]

Logs of negative numbers or zero are undefined.  As the numerator is the natural log of an <u>absolute value</u>, the numerator is undefined when:

|x - 1| = 0 ⇒ x = 1.  

A rational function is undefined when the denominator is equal to zero, so the function f(x) is undefined when x = 0.

So the function is discontinuous at x = 0 or x = 1 on the interval [-2, 2].

<u>Part (c)</u>

x = 1 is a <u>vertical asymptote</u>.  As the function exists on both sides of this vertical asymptote, it is an <u>infinite discontinuity</u>.  Since the function doesn't approach a particular finite value, the limit does not exist.  Therefore, x = 1 is a non-removable discontinuity.

A <u>hole</u> exists on the graph of a rational function at any input value that causes both the numerator and denominator of the function to be equal.

\implies f(0)=\dfrac{\ln |0-1|}{0}=\dfrac{\ln 1}{0}=\dfrac{0}{0}

Therefore, there is a hole at x = 0.

The removable discontinuity of a function occurs at a point where the graph of a function has a hole in it.   Therefore, x = 0 is a removable discontinuity.

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Answer:

  1. 109°, obtuse
  2. 131°, obtuse
  3. 53°, acute
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Step-by-step explanation:

You are exected to know the relationships of angles created where a transversal crosses parallel lines.

  • Corresponding angles are equal (congruent).
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The appearance of the diagram often gives you a clue.

You also expected to know the name (or category) of angles less than, equal to, or greater than 90°. Respectively, these are <em>acute</em>, <em>right</em>, and <em>obtuse</em> angles.

1. Adjacent angles are supplementary. The supplement of the given angle is 109°, so x will be obtuse.

2. Opposite exterior angles are equal, so y will be 131°. It is obtuse.

3. Opposite interior angles are equal, so w will be 53°. It is acute.

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We are given equation 2x+3y=9xy.

A linear equation is a equation that has maximum degree of the equation as 1.

Degree is the maximum power(exponent) of the variables.

If two variables are being multiplied together in a term, we can find power of the that term by adding powers of those variables.

In the given equation, on the left side we have two terms 2x and 3y, each opf the variable x, and y has power 1 and right side of the eqaution we have term 9xy. There x and y variables are being multiplied together.

So, the total power of the term would be 1+1=2.

So, the degree of the given equation would be 2.

Because degree of the given equation is not 1. Therefore, given equation is not a linear eqaution and it is a non-linear equation.

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