Question
Construct a parallelogram ABCD in which AB = 5.2cm, BC = 4.7cm and AC = 7.6cm.
✓
Answer
Steps of construction:Step 1: Draw AB = 5.2cm.
Step 2: With B as the centre, draw an arc of 4.7cm.
Step 3: With A as the centre, draw another arc of 7.6cm, cutting the previous arc at C.
Step 4: Join A and C.
Step 5: We know that the opposite sides of a parallelogram are equal. Thus, with C as the centre, draw an arc of 5.2cm.
Step 6: With A as the centre, draw another arc of 4.7cm, cutting the previous arc at D.
Step 7: Join CD and AD.
Then, ABCD is the required parallelogram.
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Take away:
$\frac{6}{5}\text{x}^{2}-\frac{4}{5}\text{x}^3+\frac{5}{6}+\frac{3}{2}\text{x}$ from $\frac{\text{x}^3}{3}−\frac{5}{2}\text{x}^2+\frac{3}{5}\text{x}+\frac{1}{4}$
→$\frac{6}{5}\text{x}^{2}-\frac{4}{5}\text{x}^3+\frac{5}{6}+\frac{3}{2}\text{x}$ from $\frac{\text{x}^3}{3}−\frac{5}{2}\text{x}^2+\frac{3}{5}\text{x}+\frac{1}{4}$
Solve:
$\frac{\text{2x}+7}{5}-\frac{\text{3x}+11}{2}=\frac{2\text{x}+8}{3}-5$
→$\frac{\text{2x}+7}{5}-\frac{\text{3x}+11}{2}=\frac{2\text{x}+8}{3}-5$
Verify the following:
$\frac{-13}{24}\times\Big(\frac{-12}{5}\times\frac{35}{36}\Big)=\Big(\frac{-13}{24}\times\frac{-12}{5}\Big)\times\frac{35}{36}$
→$\frac{-13}{24}\times\Big(\frac{-12}{5}\times\frac{35}{36}\Big)=\Big(\frac{-13}{24}\times\frac{-12}{5}\Big)\times\frac{35}{36}$
Solve:
$\frac{3(\text{y}−5)}{4}−4\text{y}=3-\frac{(\text{y}-3)}{2}$
→$\frac{3(\text{y}−5)}{4}−4\text{y}=3-\frac{(\text{y}-3)}{2}$