The volume of prism and other objects is important to know insdie of school and out! We learn about this stuff in school, but how can we relate it to real life? Easy! People need to know volume and area in their everyday lives, for example scientist or construction workers, and even enginneers!! so stay tuned for more info!
https://www.youtube.com/watch?v=xE3ZKkaqdUA
https://www.youtube.com/watch?v=eU2mT2hlnsY

The objective is to find the volume of a prism, pyramid, cylinder, cone, and sphere

 

 Quiz:

1.    Find the surface area of a sphere with radius 8 in. Leave your answer in terms of π.

 

 

 

2.    Find the volume of a sphere with diameter 5 in. Leave your answer in terms of π.

 

 

 

 

 

3.    If the ratio of the radii of two spheres is 3:2, what is the ratio of the surface areas of the two spheres?

 

Day 6 the objective is to learn about special right triangles

One of the most famous mathematicians who has ever lived, Pythagoras, a Greek scholar who lived way back in the 6th century B.C. (back when Bob Dole was learning geometry), came up with one of the most famous theorems ever, the Pythagorean Theorem.  It says - in a right triangle, the square of the measure of the hypotenuse equals the sum of the squares of the measures of the two legs.  This theorem is normally represented by the following equation: a² + b² = c², where c represents the hypotenuse.

With this theorem, if you are given the measures of two sides of a triangle, you can easily find the measure of the other side.
Example

1. Problem: Find the value of c.


  
  Solution: a2 + b2 = c2   Write the Pythagorean
                          Theorem and then plug in any 
                          given information.
              
            52 + 122 = c2  The information that was
                          given in the figure was
                          plugged in.
                          
                          
            169 = c2      Solve for c
            c = 13

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  45-45-90 One of the special right triangles which we deal with in geometry is an isosceles right triangle.  These triangles are also known as 45-45-90 triangles (so named because of the measures of their angles).  There is one theorem that applies to these triangles.  It is stated below.

In a 45-45-90 triangle, the measure of the hypotenuse is equal to the measure of a leg multiplied by SQRT(2).

The following figure presents the theorem in graphical terms. Ba  

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  30-60-90 Triangles   There's another kind of special right triangle which we deal with all the time.  These triangles are known as 30-60-90 triangles (so named because of the measures of their angles).  There is one theorem that applies to these triangles.  It is stated below.

In a 30-60-90 triangle, the measure of the hypotenuse is two times that of the leg opposite the 30^o angle.  The measure of the other leg is SQRT(3) times that of the leg opposite the 30^o angle.

The following figure presents the theorem in graphical terms.     

---

  Trig Ratios  While the word trigonometry strikes fear into the hearts of many, we made it through (amazing as it may seem to us), and hope to help you through it, too!  Each of the three basic trigonometric ratios are shown below. sine of angle A = (measure of opposite leg)/(measure of hypotenuse).  In the figure, the sin of angle A = (a/c).

cosine of angle A = (measure of adjacent leg)/(measure of hypotenuse).  In the figure, the cos of angle A = (b/c).

tangent of angle A = (measure of opposite leg)/(measure of adjacent leg).  In the figure, the tan of angle A = (a/b).

      1. Problem: Find sin A, cos A, and tan A.
      
      
        
        Solution: sine = (opposite/hypotenuse)
                  sine = 5/13
                  
                  
                  cosine = (adjacent/hypotenuse)
                  cos = 12/13
                  
                  tangent = (opposite/adjacent)
                  tan = 5/12

Be aware that, although the example above seems to indicate otherwise, the values for the trigonometric ratios depend on the measure of the angle, not the measures of the triangle's sides.       

---

  Story Problems  Many problems ask that you find the measure of an angle or a segment that cannot easily be measured.  Problems of this kind can often be solved by the application of trigonometry.  Below is an example problem of this type.

1. Problem: A ladder 12 meters long leans
            against a building.  It rests on
            the wall at a point 10 meters
            above the ground.  Find the angle
            the ladder makes with the ground.
            
  Solution: Make sure you know what is being
            asked.  Then use the given
            information to draw and label a
            figure.  Here's our idea of a
            figure for this problem:



            Choose a variable to represent the
            measure of the angle you are asked
            to find.  Using the variable you
            have chosen, write an equation that
            will solve the problem.
            
            sin x2 = (10/12)
            
            The above equation is derived from
            the given information and the
            knowledge of the sine
            ratio.
            
            Find the solution using a calculator's
            Arcsine function or a table
            of trigonometric ratios.
            
            TI-82 screen: sin-1 (10/12) = 56.44
            
            Trigonometric Ratios Table:
            sin 56o = 0.8290
            sin 57o = 0.8387
            
            By either answer, after rounding to
            the nearest degree, the answer is 56o.

Youtube Video:
http://www.youtube.com/watch?v=VVKUOdyzQ98

The objective of day 5 is to learn how to add and subtract radicals.

Just as with "regular" numbers, square roots can be added together. But you might not be able to simplify the addition all the way down to one number. Just as "you can't add apples and oranges", so also you cannot combine "unlike" radicals. To add radical terms together, they have to have the same radical part.

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• Simplify: 
Since the radical is the same in each term (namely, the square root of three), I can combine the terms. I have two copies of the radical, added to another three copies. This gives me five copies:

That middle step, with the parentheses, shows the reasoning that justifies the final answer. You probably won't ever need to "show" this step, but it's what should be going through your mind.

• Simplify: 
The radical part is the same in each term, so we can do this addition. To help me keep track that the first term means "one copy of the square root of three", I'll insert the "understood" "1":

Don't assume that expressions with unlike radicals cannot be simplified. It is possible that, after simplifying the radicals, the expression can indeed be simplified.

• Simplify: 
To simplify a radical addition, I must first see if I can simplify each radical term. In this particular case, the square roots simplify "completely" (that is, down to whole numbers):

• Simplify:  

I have three copies of the radical, plus another two copies, giving me— Wait a minute! I can simplify those radicals right down to whole numbers:

Don't worry if you don't see a simplification right away. If I hadn't noticed until the end that the radical simplified, my steps would have been different, but my final answer would have been the same:

• Simplify:  
I can only combine the "like" radicals, so I'll end up with two terms in my answer:

There is not, to my knowledge, any preferred ordering of terms in this sort of expression, so the expression  should also be an acceptable answer.

• Simplify:     Copyright © Elizabeth Stapel 1999-2011 All Rights Reserved
I can simplify the radical in the first term, and this will create "like" terms:

• Simplify:  I can simplify most of the radicals, and this will allow for at least a little simplification:
• Simplify:  
These two terms have "unlike" radical parts, and I can't take anything out of either radical. Then I can't simplify the expression any further and my answer has to be:


(expression is already fully simplified)
• Expand:  
To expand (that is, to multiply out and simplify) this expression, I first need to take the square root of two through the parentheses:

As you can see, the simplification involved turning a product of radicals into one radical containing the value of the product (being 2×3 = 6). You should expect to need to manipulate radical products in both "directions".

• Expand:  

• Expand:  
It will probably be simpler to do this multiplication "vertically".

Simplifying gives me:  

By doing the multiplication vertically, I could better keep track of my steps. You should use whatever multiplication method works best for you.

Quiz.
1.
Answer: Since the radicals are the same, simply add the numbers in front of the radicals (do NOT add the numbers under the radicals). 

2. Since the radicals are the same, simply add the numbers in front of the radicals (do NOT add the numbers under the radicals).

3.

1. Simplify		Answer:
2. Simplify		Answer:
3. Since the radicals in steps 1 and 2 are     now the same, we can combine them.				Answer:
4. After combining, you are left with:				Answer:
5. Can you combine these radicals?				Answer:    NO
6. Therefore,		Answer:   Youtube video: http://www.youtube.com/watch?v=MKwxPbITcXQ