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3 years ago

Write the equation of the line perpendicular to x+2y=6 that passes through (5, 8).

Mathematics

2 answers:

Bingel [31]3 years ago

6 0

Answer:

y = 2x - 2

Step-by-step explanation:

x + 2y = 6 can be solved for y (and thus for slope m) as follows:

2y = -x + 6, or y = (-1/2)x + 3.

Any new line perpendicular to this one has a slope that is the negative reciprocal of (-1/2); that would be +2.

Starting from y = mx + b,

Replace x with the given 5, y with the given 8 and m with the calculated +2:

8 = 2(5) + b, or

b = -2

and then the desired equation is

y = 2x - 2

maria [59]3 years ago

4 0

Answer:

y = 2x-2

Step-by-step explanation:

First find the slope of the line

x+2y=6

2y = -x+6

y = -1/2x +3

This is in slope intercept form ( y = mx+b) so -1/2 is the slope

We want a line that is perpendicular so take the negative reciprocal of the slope

- ( 1/ (-1/2)) = 2

The slope of the perpendicular line is 2

y = 2x+b

Substitute the point into the equation

8 = 2(5) +b

8 = 10+b

-2 =b

The equation of the line perpendicular is

y = 2x-2

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1 calcule o valor das expressões A) 2x+(3.4)²+(7.7) B) 3x+(2.6)+(8.8)² C)7.6+(28.3)+(4)² D)7x-(7.8)+(9:3)

I am Lyosha [343]

Answer:

Follows are the solution to the given points:

Step-by-step explanation:

Given:

$A) \ \ 2x+(3.4)^2+(7.7) \\\\B) \ \ 3x+(2.6)+(8.8)^2\\\\C)\ \ 7.6+(28.3)+(4)^2 \\\\D)\ \ 7x-(7.8)+(9:3)$

For point A:

$\to 2x+(3.4)^2+(7.7)\\\\\to 2x+(11.56)+(7.7)\\\\\to 2x+19.26\\\\$

For point B:

$\to 3x+(2.6)+(8.8)^2\\\\\to 3x+(2.6)+77.44\\\\\to 3x+80.04$

For point C:

$\to \ 7.6+(28.3)+(4)^2 \\\\\to \ 7.6+(28.3)+16 \\\\\to \ 7.6+44.3 \\\\\to 51.9$

For point D:

$\to 7x-(7.8)+(9:3)\\\\\to 7x-(7.8)+ \frac{9}{3}\\\\\to 7x- \frac{23.4+9}{3}\\\\\to 7x- \frac{32.4}{3}\\\\\to 7x- 10.8$

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3 years ago

How do i do 3 part a ?

WINSTONCH [101]

In general the binomial expansion is

$(a+b)^n = {n \choose 0} a^0 b^n + {n \choose 1} a^1 b^{n-1} + {n \choose 2} a^2 b^{n-2} + ... + {n \choose n} a^n b^0$

So in our case, because we want ascending powers of x we'll write,

$(-3x + 1)^{11} = {11 \choose 0} (-3x)^0 1^{11} + {11 \choose 1} (-3x)^{1} 1^{10} + {11 \choose 2} (-3x)^{2} 1^9 + {11 \choose 3 } (-3x)^3 1^8 + ...$

We need to calculate the binomial coefficients:

${11 \choose 0} = 1$

${11 \choose 1} = 11$

${11 \choose 2} = \dfrac{11 \times 10}{2} = 55$

${11 \choose 3} = \dfrac{11 \times 10 \times 9}{3 \times 2} = 165$

$(-3x+1)^{11} = 1 (-3x)^0 1^{11} + 11(-3x)^{1} 1^{10} + 55 (-3x)^2 1^{9} + 165 (-3x)^3 1^8 + ...$

$(1-3x)^{11} = 1 -33 x + 495 3x^2 - 4455 x^3+ ...$

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Answer:

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Step-by-step explanation:

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