Algebra 2
Algebra 2
1st Edition
ISBN: 9780078884825
Author: McGraw-Hill/Glencoe
Publisher: Glencoe/McGraw-Hill School Pub Co
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Concept explainers

Unitary Method
The word “unitary” comes from the word “unit”, which means a single and complete entity. In this method, we find the value of a unit product from the given number of products, and then we solve for the other number of products.
Speed, Time, and Distance
Imagine you and 3 of your friends are planning to go to the playground at 6 in the evening. Your house is one mile away from the playground and one of your friends named Jim must start at 5 pm to reach the playground by walk. The other two friends are 3 …
Profit and Loss
The amount earned or lost on the sale of one or more items is referred to as the profit or loss on that item.
Units and Measurements
Measurements and comparisons are the foundation of science and engineering. We, therefore, need rules that tell us how things are measured and compared. For these measurements and comparisons, we perform certain experiments, and we will need the experime…
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Question
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Chapter 10.1, Problem 9CYU
To determine

To find:The correct option for the distance the kiosk from the eyeglass store.

Expert Solution & Answer
Check Mark

Answer to Problem 9CYU

The correct option is A.

Explanation of Solution

Given:

The given coordinates are:

The ice cream store is at (2,4).

The eyeglass store is at (78,46).

As the kiosk for the cell phone is halfway between the ice creamery and the see clearly eyeglass.

So kiosk is at the midpoint of given coordinates first, it is need to find the midpoint M(x,y).

The midpoint of two points is given as:

  M(x,y)=(x1+x22,y1+y22)

The midpoint is calculated as:

  M(x,y)=(x1+x22,y1+y22)=(2+(78)2,4+(46)2)=(40,25)

Therefore the midpoint is (40,25) .

Now it is need to find distance between the points (40,25)&(78,46).

Now the distance between each pair of points with the given coordinates is given as:

  d=(x2−x1)2+(y2−y1)2

The distance the kiosk from the eyeglass store is calculated as:

  d=(78−40)2+(46−25)2=(1444)2+(441)2=1882≈43.38

Therefore the approximately distance the kiosk from the eyeglass store is 43.4 units.

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Chapter 10.1, Problem 8CYU
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Chapter 10.1, Problem 10PPS

Chapter 10 Solutions

Algebra 2

Ch. 10.1 - Prob. 1ACYPCh. 10.1 - Prob. 1BCYPCh. 10.1 - Prob. 2CYPCh. 10.1 - Prob. 3CYPCh. 10.1 - Prob. 1CYUCh. 10.1 - Prob. 2CYUCh. 10.1 - Prob. 3CYUCh. 10.1 - Prob. 4CYUCh. 10.1 - Prob. 5CYUCh. 10.1 - Prob. 6CYU
Ch. 10.1 - Prob. 7CYUCh. 10.1 - Prob. 8CYUCh. 10.1 - Prob. 9CYUCh. 10.1 - Prob. 10PPSCh. 10.1 - Prob. 11PPSCh. 10.1 - Prob. 12PPSCh. 10.1 - Prob. 13PPSCh. 10.1 - Prob. 14PPSCh. 10.1 - Prob. 15PPSCh. 10.1 - Prob. 16PPSCh. 10.1 - Prob. 17PPSCh. 10.1 - Prob. 18PPSCh. 10.1 - Prob. 19PPSCh. 10.1 - Prob. 20PPSCh. 10.1 - Prob. 21PPSCh. 10.1 - Prob. 22PPSCh. 10.1 - Prob. 23PPSCh. 10.1 - Prob. 24PPSCh. 10.1 - Prob. 25PPSCh. 10.1 - Prob. 26PPSCh. 10.1 - Prob. 27PPSCh. 10.1 - Prob. 28PPSCh. 10.1 - Prob. 29PPSCh. 10.1 - Prob. 30PPSCh. 10.1 - Prob. 31PPSCh. 10.1 - Prob. 32PPSCh. 10.1 - Prob. 33PPSCh. 10.1 - Prob. 34PPSCh. 10.1 - Prob. 35PPSCh. 10.1 - Prob. 36PPSCh. 10.1 - Prob. 37PPSCh. 10.1 - Prob. 38PPSCh. 10.1 - Prob. 39PPSCh. 10.1 - Prob. 40PPSCh. 10.1 - Prob. 41PPSCh. 10.1 - Prob. 42HPCh. 10.1 - Prob. 43HPCh. 10.1 - Prob. 44HPCh. 10.1 - Prob. 45HPCh. 10.1 - Prob. 46HPCh. 10.1 - Prob. 47STPCh. 10.1 - Prob. 48STPCh. 10.1 - Prob. 49STPCh. 10.1 - Prob. 50STPCh. 10.1 - Prob. 51SRCh. 10.1 - Prob. 52SRCh. 10.1 - 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Prob. 6CYUCh. 10.3 - Prob. 7CYUCh. 10.3 - Prob. 8CYUCh. 10.3 - Prob. 9CYUCh. 10.3 - Prob. 10CYUCh. 10.3 - Prob. 11CYUCh. 10.3 - Prob. 12PPSCh. 10.3 - Prob. 13PPSCh. 10.3 - Prob. 14PPSCh. 10.3 - Prob. 15PPSCh. 10.3 - Prob. 16PPSCh. 10.3 - Prob. 17PPSCh. 10.3 - Prob. 18PPSCh. 10.3 - Prob. 19PPSCh. 10.3 - Prob. 20PPSCh. 10.3 - Prob. 21PPSCh. 10.3 - Prob. 22PPSCh. 10.3 - Prob. 23PPSCh. 10.3 - Prob. 24PPSCh. 10.3 - Prob. 25PPSCh. 10.3 - Prob. 26PPSCh. 10.3 - Prob. 27PPSCh. 10.3 - Prob. 28PPSCh. 10.3 - Prob. 29PPSCh. 10.3 - Prob. 30PPSCh. 10.3 - Prob. 31PPSCh. 10.3 - Prob. 32PPSCh. 10.3 - Prob. 33PPSCh. 10.3 - Prob. 34PPSCh. 10.3 - Prob. 35PPSCh. 10.3 - Prob. 36PPSCh. 10.3 - Prob. 37PPSCh. 10.3 - Prob. 38PPSCh. 10.3 - Prob. 39PPSCh. 10.3 - Prob. 40PPSCh. 10.3 - Prob. 41PPSCh. 10.3 - Prob. 42PPSCh. 10.3 - Prob. 43PPSCh. 10.3 - Prob. 44PPSCh. 10.3 - Prob. 45PPSCh. 10.3 - Prob. 46PPSCh. 10.3 - Prob. 47PPSCh. 10.3 - Prob. 48PPSCh. 10.3 - Prob. 49PPSCh. 10.3 - Prob. 50PPSCh. 10.3 - Prob. 51PPSCh. 10.3 - 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Prob. 8CYUCh. 10.4 - Prob. 9CYUCh. 10.4 - Prob. 10CYUCh. 10.4 - Prob. 11PPSCh. 10.4 - Prob. 12PPSCh. 10.4 - Prob. 13PPSCh. 10.4 - Prob. 14PPSCh. 10.4 - Prob. 15PPSCh. 10.4 - Prob. 16PPSCh. 10.4 - Prob. 17PPSCh. 10.4 - Prob. 18PPSCh. 10.4 - Prob. 19PPSCh. 10.4 - Prob. 20PPSCh. 10.4 - Prob. 21PPSCh. 10.4 - Prob. 22PPSCh. 10.4 - Prob. 23PPSCh. 10.4 - Prob. 24PPSCh. 10.4 - Prob. 25PPSCh. 10.4 - Prob. 26PPSCh. 10.4 - Prob. 27PPSCh. 10.4 - Prob. 28PPSCh. 10.4 - Prob. 29PPSCh. 10.4 - Prob. 30PPSCh. 10.4 - Prob. 31PPSCh. 10.4 - Prob. 32PPSCh. 10.4 - Prob. 33PPSCh. 10.4 - Prob. 34PPSCh. 10.4 - Prob. 35PPSCh. 10.4 - Prob. 36PPSCh. 10.4 - Prob. 37PPSCh. 10.4 - Prob. 38PPSCh. 10.4 - Prob. 39PPSCh. 10.4 - Prob. 40HPCh. 10.4 - Prob. 41HPCh. 10.4 - Prob. 42HPCh. 10.4 - Prob. 43HPCh. 10.4 - Prob. 44HPCh. 10.4 - Prob. 45HPCh. 10.4 - Prob. 46HPCh. 10.4 - Prob. 47STPCh. 10.4 - Prob. 48STPCh. 10.4 - Prob. 49STPCh. 10.4 - Prob. 50STPCh. 10.4 - Prob. 51SRCh. 10.4 - Prob. 52SRCh. 10.4 - Prob. 53SRCh. 10.4 - 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Prob. 10STPCh. 10 - Prob. 11STPCh. 10 - Prob. 12STPCh. 10 - Prob. 13STPCh. 10 - Prob. 14STPCh. 10 - Prob. 15STP

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