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3 years ago

Which equation represents a proportional relationship? y = 3x y=2(x+1/3) y=−3x+2 y=12x

Mathematics

2 answers:

solniwko [45]3 years ago

7 0

Answer:

two equations

y=3x

and y=12x

are proportional relation ship equations

Step-by-step explanation:

proportion equations can be defined as

if we change x the y will change in same proportion

proportional relationship equation contain (0,0) points

if we put x=o

this will give us

y=0

if we put x=0

in y=12x

it will give

y=0

put if we put x=0 in

y=3x

it will give us

y=0

hence these two equation represent proportional relationship.

madreJ [45]3 years ago

6 0

The equation y = 3x and y = 12x <span>represents a proportional relationship.
Because i</span>n a proportional relationship, the graph should pass the 0,0 coordinate.
It means if you put x=0 to the equation, the result would be y=0. The graph will go in format of y=mx and if we put x = 0 in both equations we get y = 0
and in other two equations if you put x = 0 you will not get y = 0.

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Send help lol confused

hram777 [196]

Answer:

C. 0≤x≤7

Step-by-step explanation:

It is not choice D. because it has y instead of x. It is supposed to be x because it is asking for domain.

Hope this helps!

4 0

1 year ago

Erik and Caleb were trying to solve the equation: 0=(3x+2)(x-4) Erik said, "The right-hand side is factored, so I'll use the zer

Mumz [18]

Answer:

C) Both

Step-by-step explanation:

The given equation is:

$0=(3x+2)(x-4)$

To solve the given equation, we can use the Zero Product Property according to which if the product <em>A.B = 0</em>, then either A = 0 OR B = 0.

Using this property:

$(3x+2) = 0 \Rightarrow \bold{x = -\frac{2}{3}}\\(x-4) = 0 \Rightarrow \bold{x = 4}$

So, Erik's solution strategy would work.

Now, let us discuss about Caleb's solution strategy:

Multiply $(3x+2)(x-4)$ i.e. $3x^2-12x+2x-8$ = $3x^2-10x-8$

So, the equation becomes:

$0=3x^2-10x-8$

Comparing this equation to standard quadratic equation:

$ax^2+bx+c=0$

a = 3, b = -10, c = -8

So, this can be solved using the quadratic formula.

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

$x=\dfrac{-(-10)\pm\sqrt{(-10)^2-4\times3 \times (-8)}}{2\times 3}\\x=\dfrac{-(-10)\pm\sqrt{196}}{6}\\x=\dfrac{10\pm14}{6} \\\Rightarrow x= 4, -\dfrac{2}{3}$