Question

1. seven more than the quotient of a number b and 45 is greater than 5.

\frac{\square}{45} + \square \circ 5
\frac{\square}{45} \circ \square
\square \circ \square

Explanation:

Step1: Translate the first part

The quotient of \( b \) and 45 is \( \frac{b}{45} \), seven more than that is \( \frac{b}{45}+7 \), and it's greater than 5, so the first inequality is \( \frac{b}{45}+7>5 \).

Step2: Solve the inequality (subtract 7 from both sides)

Subtract 7 from both sides of \( \frac{b}{45}+7>5 \), we get \( \frac{b}{45}+7 - 7>5 - 7 \), which simplifies to \( \frac{b}{45}>-2 \).

Step3: Solve for \( b \) (multiply both sides by 45)

Multiply both sides of \( \frac{b}{45}>-2 \) by 45 (since 45 is positive, the inequality sign remains), we get \( \frac{b}{45}\times45>-2\times45 \), so \( b>-90 \).

For the blanks:

• First blank (numerator of the fraction): \( b \)
• Second blank (the "seven more" part): \( 7 \)
• The circle (inequality sign) between them and 5: \( > \)
• Then, after subtracting 7: numerator of the fraction: \( b \), circle: \( > \), the right side: \( -2 \)
• Finally, after multiplying by 45: \( b \), circle: \( > \), the right side: \( -90 \) (but for the middle step after subtraction, it's \( \frac{b}{45}>-2 \), and then \( b>-90 \))

Answer:

First row: \( \boldsymbol{b} \), \( \boldsymbol{7} \), \( \boldsymbol{>} \)

Second row: \( \boldsymbol{b} \), \( \boldsymbol{>} \), \( \boldsymbol{-2} \)

Third row: \( \boldsymbol{b} \), \( \boldsymbol{>} \), \( \boldsymbol{-90} \) (or following the steps for the blanks as per the problem's boxes, the key is the inequality setup and solution)

(If focusing on the initial inequality setup: \( \frac{b}{45}+7>5 \), then \( \frac{b}{45}>-2 \), then \( b>-90 \))