what is the equation of a line that passes through the point (6,1) and is perpendicular to the line whose equation is y=2x-6?

2 answers:

7 0

Y = (-1/2)x - 3
This line has a slope of -1/2.
Perpendicular lines have negative reciprocal slopes. The slope of the line perpendicular to the given line is 2.

y = mx + b is the slope-intercept form of the equation of a line
We have the point (6, -4) as a point on the given line. We can use these as the values of x and y in the slope-intercept form of the equation of a line.

-4 = 2(6) + b
-4 = 12 + b
-16 = b
This tells us that the line perpendicular to the given line has a y-intercept of -16. The equation of that line is:

y = 2x - 16

alexandr402 [8]3 years ago

3 0

Answer:

y = -(1/2)x + 4

Step-by-step explanation:

Use the standard form on an equation: y = mx + b

A line that is perpendicular has a slope that is the opposite reciprocal of the other line. We also have a point (x, y) that is on the line, so our equation begins as..

1 = -(1/2)(6) + b We must solve for b ( -1/2 is the opposite reciprocal of 2)

1 = -3 + b

4 = b

so our equation is

y = -(1/2)x + 4

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Long division<br> 3x+1/6x^6+5x^5+2x^4-9x^3+7x^2-10x+2

inna [77]

$(6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2) / (3x+1)$

We divide first number from first parenthesis with first number from second parenthesis. Then the resulting number we multiply by all numbers in second parenthesis and substract from first parenthesis.

$6 x^{6}/3x = 2 x^{5} \\ \\ 6 x^{6}+5x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2 - 6 x^{6} -2 x^{5} = 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2\\$

We repeat previous steps until we run out of numbers:
$3x^{5}/3x=x^{4} \\ \\ 3x^{5}+2x^{4}-9x^{3}+7x^{2}-10x+2-3x^{5}-x^{4}= \\ \\ x^{4}-9x^{3}+7x^{2}-10x+2 \\ \\ \\ x^{4}/3x= \frac{1}{3} x^{3} \\ \\ x^{4}-9x^{3}+7x^{2}-10x+2-x^{4}- \frac{1}{3} x^{3}= \\ \\ - \frac{28}{3} x^{3}+7x^{2}-10x+2 \\ \\ \\ - \frac{28}{3} x^{3}/3x= - \frac{28}{9} x^{2} \\ \\ - \frac{28}{3} x^{3}+7x^{2}-10x+2+ \frac{28}{3} x^{3}+ \frac{28}{9} x^{2} = \\ \\ \frac{91}{9} x^{2}-10x+2$
$\frac{91}{9} x^{2}/3x=\frac{91}{27} x \\ \\ \frac{91}{9} x^{2}-10x+2-\frac{91}{9} x^{2}-\frac{91}{27} x= \\ \\ -\frac{361}{27} x+2 \\ \\ \\ -\frac{361}{27} x/3x=-\frac{361}{81} \\ \\ -\frac{361}{27} x+2+\frac{361}{27}x+\frac{361}{27}= \\ \\ \frac{415}{27}$

We are left with a number that has no x inside. This is remainder.
The final solution is sum of all these solutions and remainder:
$(2 x^{5}+x^{4}+\frac{1}{3} x^{3} - \frac{28}{9} x^{2} +\frac{91}{27} x)+(-\frac{361}{81} )$

6 0

3 years ago

Solve 15+7x4 - (11 + 6)

Colt1911 [192]

Answer:28

Step-by-step explanation:

15+7•4-15

15+28-15

43-15

5 0

3 years ago

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