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m_a_m_a [10]
3 years ago
7

Sanjay said that if a line has a slope or zero, then it never touches the x-axis. Which line proves that his statement is incorr

ect a.)x=0 b.)y=0 c.)x=1 d.) y=1
Mathematics
2 answers:
salantis [7]3 years ago
6 0

Answer:

y=0  

Step-by-step explanation:

Sanjay said that if a line has a slope of zero, then it never touches the x-axis.

when slope =0, then we will get horizontal line

the horizontal line never crosses the x axis

the equation of horizontal line is y= some number

y=0 lies on x axis. it is a line that lies on x axis . that makes the statement incorrect

coldgirl [10]3 years ago
4 0
B.) y=0, because here, the slope is zero but it rests on the x-axis thus yielding an infinite amount of touching
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What two rational expressions sum to 2x+3/x^2-5x+4
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Answer:

\frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)}

Step-by-step explanation:

Given the rational expression: \frac{2x + 3}{x^2 - 5x + 4}, to express this in simplified form, we would need to apply the concept of partial fraction.

Step 1: factorise the denominator

x^2 - 5x + 4

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Thus, we now have: \frac{2x + 3}{(x- 1)(x - 4)}

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Let,

\frac{2x + 3}{(x- 1)(x - 4)} = \frac{A}{x- 1} + \frac{B}{x - 4}

Multiply both sides by (x - 1)(x - 4)

\frac{2x + 3}{(x- 1)(x - 4)} * (x - 1)(x - 4) = (\frac{A}{x- 1} + \frac{B}{x - 4}) * (x - 1)(x - 4)

2x + 3 = A(x - 4) + B(x - 1)

Step 3:

Substituting x = 4 in 2x + 3 = A(x - 4) + B(x - 1)

2(4) + 3 = A(4 - 4) + B(4 - 1)

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\frac{11}{3} = B

B = \frac{11}{3}

Substituting x = 1 in 2x + 3 = A(x - 4) + B(x - 1)

2(1) + 3 = A(1 - 4) + B(1 - 1)

2 + 3 = A(-3) + B(0)

5 = -3A

\frac{5}{-3} = \frac{-3A}{-3}

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Step 4: Plug in the values of A and B into the original equation in step 2

\frac{2x + 3}{(x- 1)(x - 4)} = \frac{A}{x- 1} + \frac{B}{x - 4}

\frac{2x + 3}{(x- 1)(x - 4)} = \frac{-5}{3(x- 1)} + \frac{11}{3(x - 4)}

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