Math Problem Statement

Convert metric and English units for weight, volume, and temperature. Round results to specified decimal places as required.

Solution

I see that you've uploaded an image containing several mathematical conversion problems. I'll process and explain how to solve these systematically.

Key Steps for Each Section:

  1. Conversions (F & G):

    • Use appropriate metric prefixes and conversion factors:
      • 1kg=103g1 \, \text{kg} = 10^3 \, \text{g}
      • 1dag=10g1 \, \text{dag} = 10 \, \text{g}
      • 1mg=103g1 \, \text{mg} = 10^{-3} \, \text{g}, etc.
    • Round results to the nearest tenth or whole number, as required.
  2. English to Metric Conversions (H & J):

    • Use given conversion constants:
      • 1lb=0.453592kg1 \, \text{lb} = 0.453592 \, \text{kg}
      • 1oz=28.3495g1 \, \text{oz} = 28.3495 \, \text{g}
      • 1T=0.907185t1 \, \text{T} = 0.907185 \, \text{t}, etc.
    • Carry answers to three decimal places as required.
  3. Volume Conversions (I):

    • For smaller to larger volumes, use:
      • 1L=10dL=1000mL1 \, \text{L} = 10 \, \text{dL} = 1000 \, \text{mL}
    • Round results to the nearest tenth.
  4. Temperature Conversions (K):

    • Use the formula C=59(F32)C = \frac{5}{9}(F - 32) for Fahrenheit to Celsius.
    • Round to the nearest degree.

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Math Problem Analysis

Mathematical Concepts

Unit Conversion
Metric System
English System
Rounding Numbers
Temperature Conversion

Formulas

1 kg = 1000 g
1 dag = 10 g
1 mg = 0.001 g
1 lb = 0.453592 kg
1 oz = 28.3495 g
1 T = 0.907185 t
C = (5/9)(F - 32)

Theorems

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Suitable Grade Level

Grades 6-8