The length of rod PR is adjusted to 16 feet. If width PQ remains the same, what is the approximate new height QR of the scaffold? Round your answer to the nearest hundredth.

Question:

Look at the picture of a scaffold used to support construction workers. The height of the scaffold can be changed by adjusting two slanting rods, one of which, labeled PR, is shown: Part A: What is the approximate length of rod PR? Round your answer to the nearest hundredth. Explain how you found your answer stating the theorem you used. Show all your work. (5 points)

B) The length of rod PR is adjusted to 16 feet. If width PQ remains the same, what is the approximate new height QR of the scaffold? Round your answer to the nearest hundredth.

A) 16.64 Feets

B) 7.65 feets

Step-by-step explanation:

Given the following :

The question above can be solved by applying Pythagoras rule :

A^2 = B^2 + C^2

PR² = PQ² + QR²

PR² = 14² + 9²

PR² = 196 + 81

PR = √277

PR = 16.64feet

Now if PR is adjusted to 16 Feets, the New height of the scaffold will be:

QR² = PR² - PQ²

QR² = 16² - 14²

QR² = 256 - 196

QR² = 60

Take the Square root of both sides

QR = √60

QR = 7.7459666 Feets

mr. flanders’ class sold 47 doughnuts and   mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations: