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3 years ago

A line passing through which of the following pairs of coordinates represents a proportional relationship?

Mathematics

2 answers:

Hoochie [10]3 years ago

7 0

D since the y-intercept would be (0,0), which means that it starts from the origin.
(haha I misread option A and B)

sasho [114]3 years ago

6 0

Answer:

Step-by-step explanation:

A proportion relation is defined as two variables that interact one to each other, directly. Basically its definition could be

$y=kx$

This means that the set of points must have a constant proportion $k$ .

However, in this case we only have one pair of points. A specific characteristic of proportional relationship in this case is that such ratio is a whole number: ±1, ±2, ±3, ±4, ±5,... ±n.

In this case, the last pair of point fulfil this characteristic. We demonstrate that by finding the ratio, which is the slope of the linear relationship

$m=\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Where $(x_{1} ,y_{1} )$ is the first point and $(x_{2} ,y_{2} )$ is the second point. Replacing these points, we have

$m=\frac{8-6}{4-3}=\frac{2}{1}=2$

So option D has a proportional relationship with a constant ratio of 2.

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Solve for x:a(a²+b²)x²+b²x-a

m_a_m_a [10]

Answer:

x = a/(a² + b²) or x = -1/a

Step-by-step explanation:

a(a²+ b²)x² + b²x - a =0

Use the quadratic equation formula:

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} =\dfrac{-b\pm\sqrt{D}}{2a}$

1. Evaluate the discriminant D

D = b² - 4ac = b⁴ - 4a(a² + b²)(-a) = b⁴ + 4a⁴ + 4a²b² = (b² + 2a²)²

2. Solve for x

$\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-b^{2}\pm\sqrt{(b^{2}+2a^{2})^{2}}}{2a(a^{2} + b^{2})}\\\\ & = & \dfrac{-b^{2}\pm(b^{2}+ 2a^{2})}{2a(a^{2} + b^{2})}\\\\x = \dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}\\\\x =\dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\\end{array}$

$\begin{array}{rcl}x = \large \boxed{\mathbf{\dfrac{a}{a^{2} + b^{2}}}}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2b^{2}- 2a^{2}}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2(a^{2}+ b^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\large \boxed{\mathbf{-\dfrac{1}{a}}}\\\\\end{array}$

5 0

3 years ago

Fifteen students are going hiking on their spring break. They plan to travel in three vehicles—one seating 7, one seating 5, and

andriy [413]

Answer:

The students can group themselves in 360360 ways

Step-by-step explanation:

For this exercise we need to use the following equation:

$\frac{n!}{n1!*n2!*...*nk!}$

This equation give us the number of assignation of n elements in k cell, where n1, n2, ..nk are the element that are in every cell

In this case we have 15 student that need to be assign in three vehicles with an specific capacity. This vehicles would be the equivalent to cells, so we can write the equation as:

$\frac{15!}{7!*5!*3!}$