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AleksandrR [38]

3 years ago

5

PLEASE HELP ME SO I CAN GRADUATE!!!!!

1 answer:

Aleks04 [339]3 years ago

7 0

Answer:

A: -41/9

Step-by-step explanation:

Substitute the point (4, -5) and slope -1/9 into the equation y = mx + b.

-5 = -1/9(4) + b

-5 = -4/9 + b

Solve for b:

b = -41/9

Our equation: y = -1/9(x) - 41/9

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Solve 5x 2 - 7x + 2 = 0 by completing the square. What are the solutions?

Alexeev081 [22]

Answer:

$x=\frac{2}{5}$ or $x=1$

Step-by-step explanation:

The given quadratic equation is

$5x^2-7x+2=0$

Group the constant terms on the right hand side.

$5x^2-7x=0-2$

$5x^2-7x=-2$

Divide through by 5.

$x^2-\frac{7}{5}x=-\frac{2}{5}$

Add the square of half the coefficient of x., which is $(\frac{1}{2}\times- \frac{7}{5})^2=\frac{49}{100}$ to both sides of the equation.

$x^2-\frac{7}{5}x+\frac{49}{100}=-\frac{2}{5}+\frac{49}{100}$

The left hand side is now a perfect square.

$(x-\frac{7}{10})^2=\frac{9}{100}$

Take the square root of both sides;

$(x-\frac{7}{10})=\pm \sqrt{\frac{9}{100}}$

$x-\frac{7}{10}=\pm \frac{3}{10}$

$x=\frac{7}{10}\pm \frac{3}{10}$

$x=\frac{7-3}{10}$ or $x=\frac{7+3}{10}$

$x=\frac{4}{10}$ or $x=\frac{10}{10}$

$x=1$ or $x=\frac{2}{5}$

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