2 years ago

Through [0, -7/5] and perpendicular to 3x-2y=6

1 answer:

Taya2010 [7]2 years ago

8 0

Find a line going through the point
(0, -7/5) and perpendicular to the equation 3x - 2y = 6.

3x - 2y = 6

-2y = -3x + 6

y = (3/2)x - 3

The slope of the equation we seek must be the negative reciprocal of (3/2), which is -2/3.

So, m = -2/3

Use the point-slope formula. See the picture.

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Please help thank you

nikdorinn [45]

Answer:

See explanation

Step-by-step explanation:

$(x^2+9x+10)(x-2)= \\\\x^3-2x^2+9x^2-18x+10x-20= \\\\x^3+7x^2-8x-20$

Jimmy is incorrect.

Using long division, you can find that the real answer.

First, subtract $x^2(x-2)$ from the original expression, leaving you with $9x^2+2x-40$ .

Next, subtract $9x(x-2)$ from the expression, leaving you with $20x-40$ . Finally, subtract $20(x-2)$ from the expression, leaving you with a remainder of 0. This means that the real quotient is $x^2+9x+20$ . Hope this helps!