X -10 -3 4 11 y 1 6 30 120
Is the relationship linear, exponential, or neither?
Choose 1
Choose 1

A
Linear

B
Exponential

c
Neither

Option C, neither.

Step-by-step explanation:

Here we have the table:

x: -10    -3    4    11

y:   1      6    30   120

This means that if our function is f(x), then:

f(-10) = 1

f(-3) = 6

f(4) = 30

f(11) = 120

We want to know if this represents a linear equation or an exponential equation or neither.

First, let's try with a linear equation.

We know that a linear equation can be written as:

f(x) = a*x + b

Then let's input the known values and let's see if this equation works.

We can use two of the known points to get:

f(-10) = 1 = a*-10 + b

f(4) = 30 = a*4 + b

With these equations, we can find the value of a and b, once we find these values, we can see if the equation also works for the other two points.

1 = a*-10 + b

30 = a*4 + b

First, we need to isolate one of the variables in one of the equations.

I will isolate b in the first one:

b = 1 + a*10

Now we can replace this in the other one to get:

30 = a*4 + 1 + a*10

30 = 1 + a*14

30 - 1 = a*14

29 = a*14

29/14 = a = 2.07

and using the equation b = 1 + a*10 we can find the value of b:

b = 1 + 2.07*10 = 1 + 20.7 = 21.7

Then the equation we get is:

f(x) = 2.07*x +  21.7

Now we need to see if this works for the other two points:

for x = -3, we need to get: f(-3) = 6

f(-3) = 2.07*-3 + 21.7 = 15.49

We did not get the value we expected, then we already know that the relationship is not linear.

Now let's see if the relationship can be exponential.

An exponential function is written as:

f(x) = A*(r)^x

Let's do the same as above, let's use two of the known points to find the values of A and r

f(-3) = 6 = A*(r)^(-3)

f(4) = 30 = A*(r)^4

Now we have the system of equations:

30 = A*(r)^4

6 = A*(r)^(-3)

If we take the quotient of these two equations, we get:

(30/6) = (A*(r)^4)/( A*(r)^(-3))

5 = (r^4)*r^3 = r^(4 + 3) = r^7

(5)^(1/7) = r = 1.258

And the value of A is given by:

30 = A*(1.258)^4

30/( (1.258)^4 ) = 11.98

Then the exponential equation is something like:

f(x) = 11.98*(1.258)^x

Now let's see if this equation also works for the other two points:

for x = -10, we should get f(-10) = 1

Let's see that:

f(-10) = 11.98*(1.258)^(-10) = 1.2

And for x = 11 we should get f(11) = 120

f(11) =  11.98*(1.258)^(11) = 149.6

So we get values closer to the ones we should get, but not the exact ones, so this is not an exponential relation.

Then the correct option is C, neither.

x = 3 ± √(5+y)

answer: the four apples you took away.

step-by-step explanation: