Answer :
Let's solve each equation step-by-step:
### Part (a)
[tex]\[ 1 = \frac{\square}{9} \][/tex]
We need to solve for the blank (let's denote it as [tex]\( x \)[/tex]):
[tex]\[ 1 = \frac{x}{9} \][/tex]
Multiply both sides by 9 to isolate [tex]\( x \)[/tex]:
[tex]\[ x = 1 \times 9 \][/tex]
[tex]\[ x = 9 \][/tex]
So, the blank in part (a) should be 9.
### Part (b)
[tex]\[ 1 = \frac{20}{\square} \][/tex]
We need to solve for the blank (let's denote it as [tex]\( y \)[/tex]):
[tex]\[ 1 = \frac{20}{y} \][/tex]
Multiply both sides by [tex]\( y \)[/tex] to isolate [tex]\( y \)[/tex]:
[tex]\[ y = 20 \][/tex]
So, the blank in part (b) should be 20.
### Part (c)
[tex]\[ \frac{2}{5} + \frac{\square}{5} = 1 \][/tex]
We need to solve for the blank (let's denote it as [tex]\( z \)[/tex]):
[tex]\[ \frac{2}{5} + \frac{z}{5} = 1 \][/tex]
Multiply both sides by 5 to clear the denominators:
[tex]\[ 2 + z = 5 \][/tex]
Subtract 2 from both sides to isolate [tex]\( z \)[/tex]:
[tex]\[ z = 5 - 2 \][/tex]
[tex]\[ z = 3 \][/tex]
So, the blank in part (c) should be 3.
### Part (d)
[tex]\[ \frac{3}{\square} + \frac{4}{7} = 1 \][/tex]
We need to solve for the blank (let's denote it as [tex]\( w \)[/tex]):
[tex]\[ \frac{3}{w} + \frac{4}{7} = 1 \][/tex]
Subtract [tex]\( \frac{4}{7} \)[/tex] from both sides:
[tex]\[ \frac{3}{w} = 1 - \frac{4}{7} \][/tex]
Convert 1 to a fraction with a denominator of 7:
[tex]\[ \frac{3}{w} = \frac{7}{7} - \frac{4}{7} \][/tex]
Subtract the fractions:
[tex]\[ \frac{3}{w} = \frac{3}{7} \][/tex]
Set the numerators equal since the denominators are the same:
[tex]\[ 3 = \frac{3w}{7} \][/tex]
Multiply both sides by 7 to isolate [tex]\( w \)[/tex]:
[tex]\[ 21 = 3w \][/tex]
Divide both sides by 3:
[tex]\[ w = \frac{21}{3} \][/tex]
[tex]\[ w = 7 \][/tex]
So, the blank in part (d) should be 7.
### Part (e)
[tex]\[ \frac{11}{\square} + \frac{\square}{20} = 1 \][/tex]
We need to solve for the blank (let's denote it as [tex]\( u \)[/tex]):
[tex]\[ \frac{11}{u} + \frac{u}{20} = 1 \][/tex]
Clear the fractions by multiplying every term by [tex]\( 20u \)[/tex]:
[tex]\[ 20u \cdot \frac{11}{u} + 20u \cdot \frac{u}{20} = 20u \cdot 1 \][/tex]
[tex]\[ 20 \cdot 11 + u^2 = 20u \][/tex]
[tex]\[ 220 + u^2 = 20u \][/tex]
Rearrange this into a standard quadratic equation:
[tex]\[ u^2 - 20u + 220 = 0 \][/tex]
Solve this quadratic using the quadratic formula [tex]\( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -20 \)[/tex], and [tex]\( c = 220 \)[/tex],
[tex]\[ u = \frac{20 \pm \sqrt{20^2 - 4 \cdot 1 \cdot 220}}{2 \cdot 1} \][/tex]
[tex]\[ u = \frac{20 \pm \sqrt{400 - 880}}{2} \][/tex]
[tex]\[ u = \frac{20 \pm \sqrt{-480}}{2} \][/tex]
Since the quadratic involves a negative under the square root, it has complex roots, but we will take the positive real part.
[tex]\[ u = \frac{20 \pm i\sqrt{480}}{2} \][/tex]
[tex]\[ u \approx 2(\frac{10 \pm i\sqrt{120}}{1}) \approx 10 + i\sqrt{120}/2 10 - i\sqrt{15}/2] Therefore we'll go with positive 20/ratio as the most straightforward root. Thus, the blank in part (e) should be 10 or un resulnant based on the quadratic) ### Part (f) \[ \frac{\square}{\square} + \frac{3}{15} = 1 \][/tex]
We need to solve for the two blanks (let's denote them as [tex]\( x \)[/tex] and [tex]\( y \)[/tex]):
[tex]\[ \frac{x}{y} + \frac{3}{15} = 1 \][/tex]
Subtract [tex]\( \frac{3}{15} \)[/tex] from both sides:
[tex]\[ \frac{x}{y} = 1 - \frac{3}{15} \][/tex]
Convert 1 to a fraction with a denominator of 15:
[tex]\[ \frac{x}{y} = \frac{15}{15} - \frac{3}{15} \][/tex]
Subtract the fractions:
[tex]\[ \frac{x}{y} = \frac{12}{15} \][/tex]
So, the ratio [tex]\( \frac{x}{y} \)[/tex] must be [tex]\( \frac{12}{15} \)[/tex]. We can choose:
[tex]\[ x = 12 \][/tex]
[tex]\[ y = 15 \][/tex]
So, the blanks in part (f) should be 12 and 15, respectively.
### Part (a)
[tex]\[ 1 = \frac{\square}{9} \][/tex]
We need to solve for the blank (let's denote it as [tex]\( x \)[/tex]):
[tex]\[ 1 = \frac{x}{9} \][/tex]
Multiply both sides by 9 to isolate [tex]\( x \)[/tex]:
[tex]\[ x = 1 \times 9 \][/tex]
[tex]\[ x = 9 \][/tex]
So, the blank in part (a) should be 9.
### Part (b)
[tex]\[ 1 = \frac{20}{\square} \][/tex]
We need to solve for the blank (let's denote it as [tex]\( y \)[/tex]):
[tex]\[ 1 = \frac{20}{y} \][/tex]
Multiply both sides by [tex]\( y \)[/tex] to isolate [tex]\( y \)[/tex]:
[tex]\[ y = 20 \][/tex]
So, the blank in part (b) should be 20.
### Part (c)
[tex]\[ \frac{2}{5} + \frac{\square}{5} = 1 \][/tex]
We need to solve for the blank (let's denote it as [tex]\( z \)[/tex]):
[tex]\[ \frac{2}{5} + \frac{z}{5} = 1 \][/tex]
Multiply both sides by 5 to clear the denominators:
[tex]\[ 2 + z = 5 \][/tex]
Subtract 2 from both sides to isolate [tex]\( z \)[/tex]:
[tex]\[ z = 5 - 2 \][/tex]
[tex]\[ z = 3 \][/tex]
So, the blank in part (c) should be 3.
### Part (d)
[tex]\[ \frac{3}{\square} + \frac{4}{7} = 1 \][/tex]
We need to solve for the blank (let's denote it as [tex]\( w \)[/tex]):
[tex]\[ \frac{3}{w} + \frac{4}{7} = 1 \][/tex]
Subtract [tex]\( \frac{4}{7} \)[/tex] from both sides:
[tex]\[ \frac{3}{w} = 1 - \frac{4}{7} \][/tex]
Convert 1 to a fraction with a denominator of 7:
[tex]\[ \frac{3}{w} = \frac{7}{7} - \frac{4}{7} \][/tex]
Subtract the fractions:
[tex]\[ \frac{3}{w} = \frac{3}{7} \][/tex]
Set the numerators equal since the denominators are the same:
[tex]\[ 3 = \frac{3w}{7} \][/tex]
Multiply both sides by 7 to isolate [tex]\( w \)[/tex]:
[tex]\[ 21 = 3w \][/tex]
Divide both sides by 3:
[tex]\[ w = \frac{21}{3} \][/tex]
[tex]\[ w = 7 \][/tex]
So, the blank in part (d) should be 7.
### Part (e)
[tex]\[ \frac{11}{\square} + \frac{\square}{20} = 1 \][/tex]
We need to solve for the blank (let's denote it as [tex]\( u \)[/tex]):
[tex]\[ \frac{11}{u} + \frac{u}{20} = 1 \][/tex]
Clear the fractions by multiplying every term by [tex]\( 20u \)[/tex]:
[tex]\[ 20u \cdot \frac{11}{u} + 20u \cdot \frac{u}{20} = 20u \cdot 1 \][/tex]
[tex]\[ 20 \cdot 11 + u^2 = 20u \][/tex]
[tex]\[ 220 + u^2 = 20u \][/tex]
Rearrange this into a standard quadratic equation:
[tex]\[ u^2 - 20u + 220 = 0 \][/tex]
Solve this quadratic using the quadratic formula [tex]\( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -20 \)[/tex], and [tex]\( c = 220 \)[/tex],
[tex]\[ u = \frac{20 \pm \sqrt{20^2 - 4 \cdot 1 \cdot 220}}{2 \cdot 1} \][/tex]
[tex]\[ u = \frac{20 \pm \sqrt{400 - 880}}{2} \][/tex]
[tex]\[ u = \frac{20 \pm \sqrt{-480}}{2} \][/tex]
Since the quadratic involves a negative under the square root, it has complex roots, but we will take the positive real part.
[tex]\[ u = \frac{20 \pm i\sqrt{480}}{2} \][/tex]
[tex]\[ u \approx 2(\frac{10 \pm i\sqrt{120}}{1}) \approx 10 + i\sqrt{120}/2 10 - i\sqrt{15}/2] Therefore we'll go with positive 20/ratio as the most straightforward root. Thus, the blank in part (e) should be 10 or un resulnant based on the quadratic) ### Part (f) \[ \frac{\square}{\square} + \frac{3}{15} = 1 \][/tex]
We need to solve for the two blanks (let's denote them as [tex]\( x \)[/tex] and [tex]\( y \)[/tex]):
[tex]\[ \frac{x}{y} + \frac{3}{15} = 1 \][/tex]
Subtract [tex]\( \frac{3}{15} \)[/tex] from both sides:
[tex]\[ \frac{x}{y} = 1 - \frac{3}{15} \][/tex]
Convert 1 to a fraction with a denominator of 15:
[tex]\[ \frac{x}{y} = \frac{15}{15} - \frac{3}{15} \][/tex]
Subtract the fractions:
[tex]\[ \frac{x}{y} = \frac{12}{15} \][/tex]
So, the ratio [tex]\( \frac{x}{y} \)[/tex] must be [tex]\( \frac{12}{15} \)[/tex]. We can choose:
[tex]\[ x = 12 \][/tex]
[tex]\[ y = 15 \][/tex]
So, the blanks in part (f) should be 12 and 15, respectively.
