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kompoz [17]
3 years ago
13

Write an equation in slope–intercept form for a line that passes through (4, 0) and is perpendicular to y = –2x.

Mathematics
1 answer:
sergeinik [125]3 years ago
7 0

Answer:


Step-by-step exp1. the equation y + 2 = -4(x - 1) needs to be in slope-intercept form. it makes it easier to solve. first, distributive property. y + 2 = -4x + 4 now you can subtract 2 from both sides to isolate y. new equation: y = -4x + 2 now you can graph this equation and the point (1, -3). we know the slope of the equation has to be the same so we can go up 4 and 1 to the left from (1, -3). this new point is (0, 1). so the answer is: y = -4x + 1


2. the graphs of the equations y = -2x + 3 and y = -2x + 7 are parallel.

i changed 2x + y = 7 to y = -2x + 7 by subtracting 2 from both sides. (standard form to slope-intercept form) then, i determined they were parallel by graphing them. you can visually see they're parallel to each other.


3. y + 1 = 2(x - 3) <- standard form. to transform this into slope-intercept form, you have to start by using the distributive property. so, do 2 times x and 2 times -3. new equation: y + 1 = 2x - 6 now you add 6 to both sides. new equation: y + 7 = 2x all you have to do now is get this in slope-intercept form. to do this, just subtract 7 from both sides. new equation: y = 2x - 7


4. the bike path is going to be parallel to main street. it is also going to pass through the park entrance (0, 4). the equation of main street is y = 2x - 3

we know that the park entrance is (0, 4) so the equation is going to look like this so far: y = x + 4 we also know the slope of the equation of main street is 2x and for the bike path to be parallel the slope needs to be the same. therefore, the equation of the line that represents the bike path is y = 2x + 4.


5. the statement: two lines with the same slope and different y-intercepts are perpendicular. is never true. this is because two lines with the same slope and different y-intercepts are parallel, not perpendicular.


i know this is a lot to read, but i like explaining things just to make sure you understand everything. the answers are in bold. i hope this helps! 



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Rufina [12.5K]

Answer:

x = \frac{1 + \sqrt{(-219)} }{22} , \space  x = \frac{1 - \sqrt{(-219)} }{22}

Step-by-step explanation:

f(x)=11x^2+x+5 is the given equation,

now comparing it with the  standard equation ax^{2} + bx + c = 0, we get

11x^{2} + x + 5 = 0

Here, a = 11, b = 1 and c = 5

Now  by QUADRATIC FORMULA

x =  \frac{-b \pm \sqrt{b^{2} - 4ac}   }{2a}

Now, b^{2}  - 4ac = 1^{2}  - 4 (11) (5) = 1 - 220 = -219

Now as discriminant D < 0, then the roots are imaginary and distinct.

So, roots are x = \frac{-1 + \sqrt{(-219)} }{22} , \space  x = \frac{-1 - \sqrt{(-219)} }{22}

These are the two roots of the given equation.

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Answer:

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Answer:

no solutions

Step-by-step explanation:

8 + 10 - 6k = 6(4 - k)

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Answer:

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Step-by-step explanation:

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