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3 years ago

What is the distance between the points (4, 7) and (4, −5) ?

Mathematics

2 answers:

Damm [24]3 years ago

5 0

Answer:

Step-by-step explanation:

distance = √(x₂-x₁)² + (y₂-y₁)²

= √(4 - 4)² + (-5 -7)²

= √0 + (-12)² = √ 144 = 12 units

pashok25 [27]3 years ago

4 0

The answer would be D, 12 units! :)

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Four decreased by the product of a number and 7 is equal to 11

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Use the Commutative Property to write an expression equivalent to –5d + 1/2.

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Answer: 12-5d

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Write an equation that is perpendicular to and 3x+y=3 whose y-intercept 5. Anyone please help me, no link just explain how to do

Liula [17]

Answer:

$y = \frac{1}{3}x + 5$

Step-by-step explanation:

By definition, two lines are perpendicular if and only if their slopes are negative reciprocals of each other: $m = - \frac{1}{m_{2} }$ , or equivalently, $m_{1} * m_{2} = -1$ .

Given our linear equation 3x + y = 3 (or y = -3x + 3):

We can find the equation of the line (with a y-intercept of 5) that is perpendicular to y = -3x + 3 by determining the negative reciprocal of its slope, -3, which is $\frac{1}{3}$ .

To test whether this is correct, we can take first slope, $m_{1} = -3$ , and multiply it with the negative reciprocal slope $m_{2} = \frac{1}{3}$ :

$m_{1} * m_{2} = -1$

$-3 * \frac{1}{3} = -1$

Therefore, we came up with the correct slope for the other line, which is $\frac{1}{3}$ .

Finally, the y-intercept is given by (0, 5). Therefore, the equation of the line that is perpendicular to 3x + y = 3 is:

$y = \frac{1}{3}x + 5$

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3 years ago

The general form of a circle is given as x^2+y^2+4x - 12y + 4 = 0. A) What are the coordinates of the center of the circle? B) W

Alex_Xolod [135]

Answer:

center = (-2, 6)

radius = 6

Step-by-step explanation:

<u>Equation of a circle</u>

$(x-a)^2+(y-b)^2=r^2$

(where (a, b) is the center and r is the radius)

Therefore, we need to rewrite the given equation into the standard form of an equation of a circle.

Given equation:

$x^2+y^2+4x-12y+4=0$

Collect like terms and subtract 4 from both sides:

$\implies x^2+4x+y^2-12y=-4$

Complete the square for both variables by adding the square of half of the coefficient of $x$ and $y$ to both sides:

$\implies x^2+4x+\left(\dfrac{4}{2}\right)^2+y^2-12y+\left(\dfrac{-12}{2}\right)^2=-4+\left(\dfrac{4}{2}\right)^2+\left(\dfrac{-12}{2}\right)^2$

$\implies x^2+4x+4+y^2-12y+36=-4+4+36$

Factor both variables:

$\implies (x+2)^2+(y-6)^2=36$

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center = (-2, 6)
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2 years ago

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