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Algebra Ch13 Test A Problem #13

Question from Grace:

We are using this test A as a practice test. We can’t figure out how to solve this.

Answer from Dr. Callahan:

Grace,

To solve this one, you take the square root of each side

+/- SQRT ( 16x^2 + 8x +1 ) = +/- SQRT (9)

The left side needs a little work to do this. Best to see if we can factor
16x^2 + 8x +1

As a hint, you know your factors would need to be the same since you are taking a square root.

So start with the first term 16x^2 – square root of it is 4x. So try that

(4x + ___) (4x + ___)

Then the last term is 1. So the obvious thing that can be multiplied by itself to get 1 is 1. So try

(4x + 1) (4x + 1)

Now when I multiply is out I get

16x^2 + 4x + 4x + 1 = 16x^2 + 8x +1

So that works!!

Now back to the problem

+/- SQRT ( 16x^2 + 8x +1 ) = +/- SQRT (9)

+/- SQRT ((4x + 1) (4x + 1)) = +/- SQRT (9)

Using the + we have
(4x + 1) = 3
And 4x = 2 so x = 1/2 is one answer

Using the – we have
(4x + 1) = -3
4x+1 = -3
4x = -4
x = -1

 

 

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Algebra Ch10.2 #3

Question from Makenna

I don’t understand how the perimeter can be 678 ft.
255 is the width x 2 = 510
505 is the length x 2 = 1010 + 510 = 1520

 

Answer from Dr. Callahan

Makenna,

You are correct. The 678 was from the older version of the book (which used a flag of 104 x 235. Flags have grown since then. 😉)

Just FYI – the older flag was flown in Detroit on the side of the J.L. Hudson Store – which no longer exists.
https://en.wikipedia.org/wiki/J._L._Hudson_Department_Store_and_Addition

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Geometry Ch11 test #7b

Question from Anya 
Course: Geometry
Chapter/Lesson/Problem11 test7b
Issue: I’m supposed to find Angle APM. To find the angle, I used tan APM=20/70, which gives 15.9 degrees. The solutions manual used sin APM=20/70, which gives 16.6 degrees. Sine is opposite/hypotenuse. However, it seems to me that the given measurements of the problem are opposite side (20) to the adjacent side (70). Which is correct? Thank you!
Answer by Dr.Callahan
Yes, you are correct. That should be tangent as you stated.
tan(APM) = 20/70
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Algebra Ch5.6 #15

Ch5.6 #15

As before – all are similar so I will pick the harder one.

Givens

AB = CD

AB = 9(x-2)

CD = x + 3(x+5)

Therefore,

9(x-2) = x + 3(x+5)

Now multiply out and solve

9x – 18 = x + 3x + 15

9x – x – 3x  = 15 + 18

5x = 33

x = 6.6

Now go back and find the lengths

AB = 9(x-2) = 9(6.6 – 2) = 41.4

CD = AB = 41.4

CE = x = 6.6

ED = 3(x+5) = 3(6.6+5) = 34.8

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Algebra Ch5.6 #14

Now let’s look at Ch 5.6 #14.

Parts a-c are similar – so will do part c since the harder.

Area = x + 20 = (x-1)4    > I am just stating all things that are equal based on the given.

x+20 = (x-1)4

x+20 = 4x -4  > multiplied out

20 + 4 = 4x – x  > moved similar items to same side

24 =  3x

x = 8

Now we know x, we have to find the sides (ALWAYS GO BACK AND MAKE SURE YOU KNOW WHAT WAS ASKED)

x -1  = 8 – 1 = 7

4 = 4

 

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Algebra Ch5.6 #9

For problem 5.6, parts a-d are all pretty much the same. Here I will do part d)  since a hair more complex. The steps the same for the others.

Since AB = CD, we can substitute the following

AB is x + 6(x+2)

CD is 2(x+10)

To get

x + 6(x+2)  = 2(x+10)

So to find the length of AB and CD, I need to figure out x. So solve the equation.

x + 6(x+2)  = 2(x+10)

x + 6x + 12 = 2x + 20   – I just multiplied everything out

x + 6x – 2x = 20 – 12  > I moved all the x terms to the left and all the constants (numbers) to the right

5x = 8  > did the math

x = 8/5  = 1.6

Now I have x, but the problem asked for the lengths.

So now we can find the parts

AB = x + 6(x+2) = 1.6 + 6(1.6 + 2) = 1.6 + 6(3.6) = 23.2

AE = x = 1.6

EB = 6(x+2) = 21.6

CD = AB = 23.2

 

 

To do the others – follow the same process

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Geometry 14.2 #22-23

Question from Joy:

We are in Geometry 3 rd edition. Chapter 14 lesson 2 number 22-23. Page 582. 
How do you get 108?

The answer in key says 3 x 180/5?  Why? I think I missed something important yet simple. I was thinking divide 360/5. What an I doing wrong?  Thanks!

Answer from Dr. Callahan:

I have to admit – I could not find this in the book. I think it SHOULD be there – but could not find it.

The answer is the a triagle has interior angles that add up to 180 (we know that – right)

So each time we add a side (triangle to square) we add 180.

See here https://www.mathsisfun.com/geometry/interior-angles-polygons.html

So in #22, he is saying we have 180 3 times (triangle plus 2 sides adding 2 more 180s) as the total angles and then dived by the number of sides to get 108.

Again – I do not see this in the book.

Hope this helps

dwc

 

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Geometry Ch13.5 #15-18

For Chap 13, Lesson 6, #32, the basis for the answers is the figure for #6.  If we look carefully at the grid, applying the Pythagorean Theorem gives XY^2 = 2^2 + 2^2 = 8, so XY = sqrt8 = 2sqrt2.  YZ^2 = 1^2 + 1^2 = 2, so YZ =sqrt2.  Therefore XY = 2YZ.
Harold Jacobs

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The Different Geometry Textbooks and Videos

The 3rd edition of the Harold Jacobs Geometry textbook has gone through three different printings with 3 different covers and ISBNs. If you are picking a text and looking at used options, it can be confusing. So let me try to help.

First, all 3 versions are basically the same – as long as they say 3rd edition. They look like the photos below.

The Three Geometry Textbook Versions that Still Work

The original published by Freeman – ISBN: 978-0-7167-4361-3 and the corresponding Teachers Guide which contains the solutions to the problems. This book had a separate Test Bank for tests.

 

 

The second printing published by My Father’s World – ISBN: 978-1-61999-109-5 and the corresponding Teachers Guide which contained the tests and the Answer Key.

 

 

The third printing published by Master Books – ISBN: 9787-1-68344-020-8. This is the latest and if you are buying new, this is what you are getting. The tests are in the Teacher’s Guide and the answers are in the Solutions Manual. You will need all 3 books for the course.

Jacobs Geometry Curriculum

 

 

 

 

 

 

 

 

 

The Geometry Videos that Support the above Books

All 3 of the printings will work with the AskDrCallahan videos. The various printings have minor changes in page numbers, drawings, and some problems have changed – but the basic content remains the same. But, just to be more confusing, the videos come in 3 different packages. All are the same content but they might look different. All require the student to have one of the 3 textbooks, the tests, and a set of solutions.

 

1 – The original DVDs published by us (AskDrCallahan) -ISBN:978-0-615-27810-0

 

 

2 – The latest version of the DVDs was published by Master Books – UPC: 713438-110236-8

 

 

 

 

 

 

 

 

 

 

 

 

3 – The online version, offered by us only, have the same content and bypass the need for DVDs and a DVD player. They are the same content as on the DVD. The online videos come in two options.

  • Monthly option – Pay for the course monthly and cancel anytime by logging into your AskDrCallahan account (or email us to cancel). This option is perfect if you think you will need just a few months.
  • Lifetime access – If you need a full year, or have siblings who might use the course later, then this is your option. One price one time.

 

 

 

 

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Geometry Ch8 Review #32

Question from Jennifer:
Chap 8 review
#32
We aren’t sure how the answer would be 24mm.
Answer:
This is not explained well in the textbook.
First – the translation is the movement of the bat from position B to position C (see chap 8, Lesson 2) A translation is two rotations around a parallel line – or in other words, we moved it over from one place to another.
So he is asking, how far did it move. Or, what is the length of the black line moving from B to C.
Since we can measure the bat to be about 42 mm (with a ruler), the length of the black line is about 24 mm using the same ruler.