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3 years ago

What is the equation of the line that is perpendicular to the given line and has an x-intercept of -3?

Mathematics

1 answer:

Anvisha [2.4K]3 years ago

3 0

Answer:

$y=-\frac{3}{2}x-\frac{9}{2}$

Step-by-step explanation:

Let the equation of the perpendicular line is,

y = mx + b

where m = slope of the line

b = y-intercept

From the graph, slope of the line passing through (0, -1) and (3, 1),

m' = $\frac{y_2-y_1}{x_2-x_1}$

m' = $\frac{1+1}{3-0}$

m' = $\frac{2}{3}$

To get the slope (m) of this line we will use the property of perpendicular lines,

m × m' = (-1)

m × $\frac{2}{3}$ = -1

m = $-\frac{3}{2}$

Equation of the perpendicular line will be,

$y=-\frac{3}{2}x+b$

x-intercept of the line is (-3) therefore, point on the line is (-3, 0)

0 = $-\frac{3}{2}(-3)+b$

b = $-\frac{9}{2}=-4.5$

Equation of the line will be,

$y=-\frac{3}{2}x-\frac{9}{2}$

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2 years ago

The sum of two consecutive even integers is at most seven more than half the sum of the next two consecutive even integers

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The sum of two consecutive even integers is at most seven more than half the sum of the next two consecutive even integers

two consecutive even integers: n, n+2

next two consecutive even integers n+4, n+6

n+(n+2) >= 7+1/2(n+4 +n+6)

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3 years ago

Part A: Factor x2y2 + 6xy2 + 8y2. Show your work. (4 points) Part B: Factor x2 + 8x + 16. Show your work. (3 points) Part C: Fac

skad [1K]

Answer:

Part A : y²(x + 2)(x + 4)

Part B: (x + 4) (x + 4)

Part C: (x + 4) (x - 4)

Step-by-step explanation:

Part A: Factor x²y²+ 6xy²+ 8y²

x²y²+ 6xy²+ 8y²

y² is very common across the quadratic equation , hence

= y² (x² + 6x + 8)

= (y²) (x² + 6x + 8)

= (y²) (x² + 2x +4x + 8)

= (y²) (x² + 2x)+(4x + 8)

= (y²) (x(x + 2)+ 4(x + 2))

= y²(x+2)(x+4)

Part B: Factor x² + 8x + 16

x² + 8x + 16

= x² + 4x + 4x + 16

= (x² + 4x) + (4x + 16)

= x( x + 4) + 4(x + 4)

= (x + 4) (x + 4)

Part C: Factor x² − 16

= x² − 16

= x² + 0x − 16

= x² + 4x - 4x - 16

= (x² + 4x) - (4x - 16)

= x (x + 4) - 4(x + 4)

= (x + 4) (x - 4)

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3 years ago

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SVEN [57.7K]

Answer:

$\boxed{\sf distance \: between \: the \: objects \: \: is \: 2.5 \: cm}$

Step-by-step explanation:

According to universal law of gravitation every object in the universe attracts every other particles that surrounds it with a force which is inversely proportional to the square of their distance of separation & directly proportional to

the product of their masses, given by standard formula.

$A = G \cdot \frac{m_1.m_2}{ {d}^{2} }$

where A,d & G are force of attraction, distance of separation & proportionality constant respectively.

Given:

A1 = 2 units

d1= 5 cm

A2 = 8 units

To find:

Distance of separation when the force of attraction is 8 units d2 = ?

Solution:

Substituting the given values in above at each point,

$A_1 = G \cdot \frac{m_1.m_2}{ {d_1}^{2} } \\ 2 = G \cdot \frac{m_1.m_2}{ {5}^{2} } \\ \sf similarly \: at \: second \: point \: of \: attraction \\ A_2 = G \cdot \frac{m_1.m_2}{ {d_2}^{2} } \\ 8 = G \cdot \frac{m_1.m_2}{ {d_2}^{2} } \\ \sf \: \: dividing \: both \: of \: the \: equation \\ \frac{2}{8} = \frac{G \cdot \frac{m_1.m_2}{ {5}^{2} }}{G \cdot \frac{m_1.m_2}{ {d_2}^{2} } } \\ \sf \: G \: m_1 and \: m_2 \: are \: \: constant \: hence \\ \sf they \: can \: be \: cancelled \: out \\ \frac{2}{8} = \frac{ \frac{1}{ {5}^{2} } }{ \frac{1}{ {d_2}^{2} } } \\ \sf rearranging \: above \: equation \\ \frac{1}{4} = \frac{{d_2}^{2} }{ {5}^{2} } \\ {d_2}^{2} = \frac{ {5}^{2} }{4 } \\ {d_2}^{2} = \frac{25}{4} \\ {d_2}^{2} = 6.25 \\ \sqrt{{d_2}^{2} } = \sqrt{6.25} \\ \boxed{ \sf{d_2} = 2.5 \: cm}$

Answer: the distance between the two objects is 2.5 cm, if the attraction between them is 8 units.

Learn more about universal law of gravitation here brainly.com/question/27244479

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