What is the equation of a line that has the same slope as y=7-2x and the same y-intercept as y=-7x-5

Mathematics

1 answer:

wariber [46]2 years ago

3 0

Answer:

Hello!

First you will want to get both of equations into slope-intercept form. That way, you can easily determine the y-intercept of the first equation and the slope of the second equation.

When 2 lines have the same slope, that means they will be parallel.

y+4x-10=0 ; add 10 to each side

y+4x=10 ; subtract 4x from each side

y = -4x + 10 ; y-intercept = 10

1x + 7y = 6 ; subtract 1x from each side

7y = -1x + 6 ; divide both sides by 7

y = -1/7 x + 6/7 ; slope = -1/7

So, a line with the same y-intercept as the first, and same slope (parallel) as the second would have the equation:

y = -1/7x + 10

Step-by-step explanation:

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barxatty [35]

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4 0

3 years ago

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calculeaza lungimea segmentului ab in fiecare dintre cazuri:A(1,5);B(4,5);A(2,-5),B(2,7);A(3,1)B(-1,4);A(-2,-5)B(3,7);A(5,4);B(-

Tatiana [17]

Answer:

1. 3; 2. 12; 3. 5; 4. 13; 5. 10; 6. 10

Step-by-step explanation:

We can use the distance formula to calculate the lengths of the line segments.

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

1. A (1,5), B (4,5) (red)

$d = \sqrt{(x_{2} - x_{1}^{2}) + (y_{2} - y_{1})^{2}} = \sqrt{(4 - 1)^{2} + (5 - 5)^{2}}\\= \sqrt{3^{2} + 0^{2}} = \sqrt{9 + 0} = \sqrt{9} = \mathbf{3}$

2. A (2,-5), B (2,7) (blue)

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(2 - 2)^{2} + (7 - (-5))^{2}}\\= \sqrt{0^{2} + 12^{2}} = \sqrt{0 + 144} = \sqrt{144} = \mathbf{12}$

3. A (3,1), B (-1,4 ) (green)

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(-1 - 3)^{2} + (4 - 1)^{2}}\\= \sqrt{(-4)^{2} + 3^{2}} = \sqrt{16 + 9} = \sqrt{25} = \mathbf{5}$

4. A (-2,-5), B (3,7) (orange)

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(3 - (-2))^{2} + (7 - (-5))^{2}}\\= \sqrt{5^{2} + 12^{2}} = \sqrt{25 + 144} = \sqrt{169} = \mathbf{13}$

5. A (5,4), B (-3,-2) (purple)

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(-3 - 5)^{2} + (-2 - 4)^{2}}\\= \sqrt{(-8)^{2} + (-6)^{2}} = \sqrt{64 + 36} = \sqrt{100} = \mathbf{10}$

6. A (1,-8), B (-5,0) (black)

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(-5 - 1)^{2} + (0 - (-8))^{2}}\\-= \sqrt{(-6)^{2} + (-8)^{2}} = \sqrt{36 + 64} = \sqrt{100} = \mathbf{10}$