Question

Pls can someone help me and have it right

Answer 1

Answer:

Problem 1:

a. x=2

b. x=3

c. x=1

Problem 2:

A multiplication equation to hold the table true:

$18\times3=54$

A division equation to hold the table true:

$\frac{54}{3} =18$

Step-by-step explanation:

Given in problem 1:

(a). The equation is $\frac{x}{6} > 1$

It holds true for all values of $x> 6$.

Let us say $x=12$,

$\frac{x}{6}=\frac{12}{6} =2$ which is greater than 1.

(b). The equation is $\frac{x}{6} < 1$

It holds true for all values of $x< 6$.

Let us say $x=3$

$\frac{x}{6} =\frac{3}{6} =\frac{1}{2}$ which is less than 1.

(c). The equation is $\frac{x}{6} = 1$

It holds true for only $x= 6$.

Let us say $x=6$,

$\frac{x}{6} =\frac{6}{6}=1$ which is equal to 1.

Problem 2:

A multiplication equation to hold the table true:

$18\times3=54$

A division equation to hold the table true:

$\frac{54}{3} =18$

Therefore these are the values which hold true to the equation in problem 1 and 2.