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Effectus [21]
3 years ago
7

Margaret is planning a wedding reception. She wants to spend less than her budget of $5,748, and Margaret has already spent $2,4

78. The meal price per guest is $30. How many guests, x, can Margaret invite to the reception and stay under budget? Select the number line that includes the largest number of guests Margaret can invite without exceeding her budget.
Mathematics
1 answer:
noname [10]3 years ago
6 0

Answer:

She can invite 109 guests and be right on top of budget. If she wants to spend less than her budget though she can only invite 108 people.

Step-by-step explanation:

First you have to subtract 5,748 - 2,478 which is 3270. Next you have to divide 3270 by 30 because each person costs 30 dollars. 3270 divided by 30 is 109.

I hope this helps! Please mark me brainliest if I am correct! Have a nice day!

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lys-0071 [83]

Answer:

ASA

ΔFGH ≅ ΔIHG ⇒ answer B

Step-by-step explanation:

* Lets revise the cases of congruence  

- SSS  ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ

- SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and  

 including angle in the 2nd Δ

- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ  

 ≅ 2 angles and the side whose joining them in the 2nd Δ

- AAS ⇒ 2 angles and one side in the first triangle ≅ 2 angles  

 and one side in the 2ndΔ  

- HL ⇒ hypotenuse leg of the first right angle triangle ≅ hypotenuse

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* Lets prove the two triangles FGH and IHG are congruent by on of

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∵ FG // HI and GH is transversal

∴ m∠FGH = m∠IHG ⇒ alternate angles

- In the two triangles FGH and IHG

∵ m∠FHG = m∠IGH ⇒ given

∵ m∠FGH = m∠IHG ⇒ proved

∵ GH = HG ⇒ common side

∴ ΔFGH ≅ ΔIHG ⇒ ASA

* ASA

 ΔFGH ≅ ΔIHG

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