Steve and Michelle Neely are celebrating their 30th anniversary by having a reception at a local reception hall. They have budgeted $4,000 for their reception. If the reception hall charges $50 cleanup charge fee plus $33 per person, find the greatest number of people that they may invite and still stay within their budget

Greatest Number of People that may be invited = 48

Step-by-step explanation:

Step 1:

Let the budget be $ 4000 =Amount

Total cost per person =( $50 + $33 ) = $83

Step 2:

Greatest number of persons Steve and Michelle Neely may invite is given by

Greatest number of  Persons may be invited

                                                  = (Budget decided) / (Cost per person)

People Count = ( 4000) / ( 83) = 48.19

                        =  (48*83) + 16 = 4000

Tentatively Greatest number of people may be invited is equal to 48 and remaining budget with them is $16

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°