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4 years ago

Combine like terms 6x - (5x) + 2x + (-3)

Mathematics

1 answer:

Leya [2.2K]4 years ago

6 0

Answer:

13x - 3

Step-by-step explanation:

Step 1:

6x - ( 5x ) + 2x + ( - 3 ) Equation

Step 2:

6x + 5x + 2x - 3 Open Parenthises

Answer:

13x - 3 Add

Hope This Helps :)

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Line l contains the points (1,5) and (4, -4). point P has the coordinates (-1, 1)

Colt1911 [192]

Answer:

The distance from p to l is $\sqrt{10}\ units$

Step-by-step explanation:

we know that

The distance between point p from line l is equal to the perpendicular segment from line l to point p

step 1

<em>Find the slope of line l</em>

we have the points

(1,5) and (4, -4)

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{-4-5}{4-1}$

$m=\frac{-9}{3}$

$m=-3$

step 2

Find the equation of the line l

The equation in point slope form is equal to

$y-y1=m(x-x1)$

we have

$m=-3$

$point\ (1,5)$

substitute

$y-5=-3(x-1)$ -----> equation A

step 3

Find the slope of the line perpendicular to the line l

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (The product of their slopes is equal to -1)

$m_1*m_2=-1$

we have

$m_1=-3$ ---> slope of line l

therefore

$m_2=\frac{1}{3}$ ----> slope of the line perpendicular to line l

step 4

Find the equation of the line perpendicular to line l that passes through the point p

The equation in point slope form is equal to

$y-y1=m(x-x1)$

we have

$m=\frac{1}{3}$

$point\ p(-1,1)$

substitute

$y-1=\frac{1}{3}(x+1)$ -----> equation B

step 5

Solve the system of equations

$y-5=-3(x-1)$ -----> equation A

$y-1=\frac{1}{3}(x+1)$ -----> equation B

Solve the system by graphing

The solution of the system is the intersection point both graphs

The solution is the point q(2,2)

see the attached figure

step 6

we know that

The distance between the point p and the line l is equal to the distance between the point p and the point q

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have the points

p(-1,1) and q(2,2)

substitute the values

$d_p_q=\sqrt{(2-1)^{2}+(2+1)^{2}}$

$d_p_q=\sqrt{(1)^{2}+(3)^{2}}$

$d_p_q=\sqrt{10}\ units$

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romanna [79]

Answer:

it just takes longer

Step-by-step explanation:

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