Answer :
Sure! Let's solve the given equations step-by-step to find the missing numbers.
### First Equation:
[tex]\[ J \times 4 = 20 \][/tex]
To find [tex]\( J \)[/tex], we divide both sides of the equation by 4:
[tex]\[ J = \frac{20}{4} = 5 \][/tex]
So, [tex]\( J = 5 \)[/tex].
### Second Equation:
[tex]\[ 15 \times X = 60 \][/tex]
To find [tex]\( X \)[/tex], we divide both sides of the equation by 15:
[tex]\[ X = \frac{60}{15} = 4 \][/tex]
So, [tex]\( X = 4 \)[/tex].
### Third Equation:
[tex]\[ Y \div 8 = 9 \][/tex]
To find [tex]\( Y \)[/tex], we multiply both sides of the equation by 8:
[tex]\[ Y = 9 \times 8 = 72 \][/tex]
So, [tex]\( Y = 72 \)[/tex].
### Fourth Equation:
[tex]\[ Z \div 4 = 3 \][/tex]
To find [tex]\( Z \)[/tex], we multiply both sides of the equation by 4:
[tex]\[ Z = 3 \times 4 = 12 \][/tex]
So, [tex]\( Z = 12 \)[/tex].
Therefore, the filled-in equations and their solutions are:
[tex]\[ \begin{pmatrix} 5 & \times & 4 & = & 20 \\ \end{pmatrix} \][/tex]
[tex]\[ \begin{pmatrix} 15 & \times & 4 & = & 60 \\ \end{pmatrix} \][/tex]
[tex]\[ \begin{pmatrix} 72 & \div & 8 & = & 9 \\ \end{pmatrix} \][/tex]
[tex]\[ \begin{pmatrix} 12 & \div & 4 & = & 3 \\ \end{pmatrix} \][/tex]
The missing numbers are [tex]\( J = 5 \)[/tex], [tex]\( X = 4 \)[/tex], [tex]\( Y = 72 \)[/tex], and [tex]\( Z = 12 \)[/tex].
### First Equation:
[tex]\[ J \times 4 = 20 \][/tex]
To find [tex]\( J \)[/tex], we divide both sides of the equation by 4:
[tex]\[ J = \frac{20}{4} = 5 \][/tex]
So, [tex]\( J = 5 \)[/tex].
### Second Equation:
[tex]\[ 15 \times X = 60 \][/tex]
To find [tex]\( X \)[/tex], we divide both sides of the equation by 15:
[tex]\[ X = \frac{60}{15} = 4 \][/tex]
So, [tex]\( X = 4 \)[/tex].
### Third Equation:
[tex]\[ Y \div 8 = 9 \][/tex]
To find [tex]\( Y \)[/tex], we multiply both sides of the equation by 8:
[tex]\[ Y = 9 \times 8 = 72 \][/tex]
So, [tex]\( Y = 72 \)[/tex].
### Fourth Equation:
[tex]\[ Z \div 4 = 3 \][/tex]
To find [tex]\( Z \)[/tex], we multiply both sides of the equation by 4:
[tex]\[ Z = 3 \times 4 = 12 \][/tex]
So, [tex]\( Z = 12 \)[/tex].
Therefore, the filled-in equations and their solutions are:
[tex]\[ \begin{pmatrix} 5 & \times & 4 & = & 20 \\ \end{pmatrix} \][/tex]
[tex]\[ \begin{pmatrix} 15 & \times & 4 & = & 60 \\ \end{pmatrix} \][/tex]
[tex]\[ \begin{pmatrix} 72 & \div & 8 & = & 9 \\ \end{pmatrix} \][/tex]
[tex]\[ \begin{pmatrix} 12 & \div & 4 & = & 3 \\ \end{pmatrix} \][/tex]
The missing numbers are [tex]\( J = 5 \)[/tex], [tex]\( X = 4 \)[/tex], [tex]\( Y = 72 \)[/tex], and [tex]\( Z = 12 \)[/tex].
