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Xelga [282]
2 years ago
13

Li believes that the graph shows a direct variation. Why is Li incorrect in saying that the graph shows a direct variation?

Mathematics
2 answers:
padilas [110]2 years ago
3 0

Answer:

Option 2) When the x-value is 0, the y-value is 1.

Step-by-step explanation:

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form y/x=k or y=kx

In a proportional relationship the constant of proportionality k is equal to the slope m of the line <em>and the line passes through the origin </em>

Remember that in a direct variation

For x=0, the value of y is equal to zero too

therefore

The graph is not a direct variation , because

When the x-value is 0, the y-value is 1.

Ugo [173]2 years ago
3 0

Answer:

b. or 2. When the x-value is 0, the y-value is 1.

Step-by-step explanation:

I just took the quiz ;)

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Integral of 17/(x^3-125)
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Answer:

17/75 ln│x − 5│− 17/150 ln(x² + 5x + 25) − 17/(5√75) tan⁻¹((2x + 5) / √75) + C

Step-by-step explanation:

∫ 17 / (x³ − 125) dx

= 17 ∫ 1 / (x³ − 125) dx

= 17 ∫ 1 / ((x − 5) (x² + 5x + 25)) dx

Use partial fraction decomposition:

= 17 ∫ [ A / (x − 5) + (Bx + C) / (x² + 5x + 25) ] dx

Use common denominator to find the missing coefficients.

A (x² + 5x + 25) + (Bx + C) (x − 5) = 1

Ax² + 5Ax + 25A + Bx² − 5Bx + Cx − 5C = 1

(A + B) x² + (5A − 5B + C) x + 25A − 5C = 1

Match the coefficients and solve the system of equations.

A + B = 0

5A − 5B + C = 0

25A − 5C = 1

A = 1/75

B = -1/75

C = -2/15

So the integral is:

= 17 ∫ [ 1/75 / (x − 5) + (-1/75 x − 2/15) / (x² + 5x + 25) ] dx

Simplify:

= 17/75 ∫ [ 1 / (x − 5) − (x + 10) / (x² + 5x + 25) ] dx

Factor ½ from the numerator of the second fraction:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 20) / (x² + 5x + 25) ] dx

Split the fraction:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − ½ (15) / (x² + 5x + 25) ] dx

Multiply the last fraction by 4/4:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − 30 / (4x² + 20x + 100) ] dx

Complete the square:

= 17/75 ∫ [ 1 / (x − 5) − ½ (2x + 5) / (x² + 5x + 25) − 15 / ((2x + 5)² + 75) ] dx

Split the integral:

= 17/75 ∫ 1 / (x − 5) dx − 17/150 ∫ (2x + 5) / (x² + 5x + 25) dx − 17/5 ∫ 1 / ((2x + 5)² + 75) dx

The first integral is:

∫ 1 / (x − 5) dx = ln│x − 5│

The second integral is:

∫ (2x + 5) / (x² + 5x + 25) dx = ln(x² + 5x + 25)

The third integral is:

∫ 1 / ((2x + 5)² + 75) dx = 1/√75 tan⁻¹((2x + 5) / √75)

Plug in:

= 17/75 ln│x − 5│− 17/150 ln(x² + 5x + 25) − 17/(5√75) tan⁻¹((2x + 5) / √75) + C

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