Jon is drawing a quadrilateral. The length of each side is 5.2cm to 1 decimal place.

Complete the error interval for the length of one side.

And Complete the error interval for the perimeter.

Please help with this one urgently.

We know that the measure of each side is:

5.2cm to 1 decimal place.

Then we can assume that the error is 0.05cm (half of the second decimal place).

With this, the interval of the length of each side will be:

(5.2cm - 0.05cm) = 5.15cm  (as the minimum)

(5.2cm + 0.05cm) = 5.25cm (as the maximum)

Then the interval is (5.15cm, 5.25cm)

2) We know that the perimeter of a square of sidelength L is:

P = 4*L

In this case, the minimum perimeter will be if we choose the smallest sidelength, this is:

P = 4*5.15cm = 20.6cm

And the maximum perimeter can be:

P = 4*5.25cm = 21cm

Then the interval is: (20.6cm, 21cm)

Here we can see that the error for the perimeter is 0.2cm, which is equal to 4 times the error of the length of a side (0.05cm) , as expected.

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°