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3 years ago

6. Find the missing OUTPUT value.

Mathematics

1 answer:

matrenka [14]3 years ago

6 0

Answer:

Which table represents a nonlinear function?

X 1

2 3 4

y 3 6 9 12 15 18

X 2 4 6

10 12

y 4 7 10 13 16 19

X -2 -1 0 1

2 3

4 2 1 1.5 0.5

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3 years ago

without building the graph, find the coordinates of the point of intersection of the lines given by the equation y=3x-1 and 3x+y

DaniilM [7]

<h2><u>1. Determining the value of x and y:</u></h2>

Given equation(s):

$y = 3x - 1$
$3x + y = -7$

To determine the point of intersection given by the two equations, it is required to know the x-value and the y-value of both equations. We can solve for the x and y variables through two methods.

<h3 /><h3><u>Method-1: Substitution method</u></h3>

Given value of the y-variable: 3x - 1

Substitute the given value of the y-variable into the second equation to determine the value of the x-variable.

$\implies 3x + y = -7$

$\implies3x + (3x - 1) = -7$

$\implies3x + 3x - 1 = -7$

Combine like terms as needed;

$\implies 3x + 3x - 1 = -7$

$\implies 6x - 1 = -7$

Add 1 to both sides of the equation;

$\implies 6x - 1 + 1 = -7 + 1$

$\implies 6x = -6$

Divide 6 to both sides of the equation;

$\implies \dfrac{6x}{6} = \dfrac{-6}{6}$

$\implies x = -1$

Now, substitute the value of the x-variable into the expression that is equivalent to the y-variable.

$\implies y = 3(-1) - 1$

$\implies$ $\ \ = -3 - 1$

$\implies$ $= -4$

Therefore, the value(s) of the x-variable and the y-variable are;

$\boxed{x = -1}$ $\boxed{y = -4}$

<h3 /><h3><u>Method 2: System of equations</u></h3>

Convert the equations into slope intercept form;

$\implies\left \{ {{y = 3x - 1} \atop {3x + y = -7}} \right.$

$\implies \left \{ {{y = 3x - 1} \atop {y = -3x - 7}} \right.$

Clearly, we can see that "y" is isolated in both equations. Therefore, we can subtract the second equation from the first equation.

$\implies \left \{ {{y = 3x - 1 } \atop {- (y = -3x - 7)}} \right.$

$\implies \left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.$

Now, we can cancel the "y-variable" as y - y is 0 and combine the equations into one equation by adding 3x to 3x and 7 to -1.

$\implies\left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.$

$\implies 0 = (6x) + (6)$

$\implies0 = 6x + 6$

This problem is now an algebraic problem. Isolate "x" to determine its value.

$\implies 0 - 6 = 6x + 6 - 6$

$\implies -6 = 6x$

$\implies -1 = x$

Like done in method 1, substitute the value of x into the first equation to determine the value of y.

$\implies y = 3(-1) - 1$

$\implies y = -3 - 1$

$\implies y = -4$

Therefore, the value(s) of the x-variable and the y-variable are;

$\boxed{x = -1}$ $\boxed{y = -4}$

<h2><u>2. Determining the intersection point;</u></h2>

The point on a coordinate plane is expressed as (x, y). Simply substitute the values of x and y to determine the intersection point given by the equations.

⇒ (x, y) ⇒ (-1, -4)

Therefore, the point of intersection is (-1, -4).

<h3>Graph:</h3>