Friday, May 24, 2013

Unit 2: Day 8

Today, students will use the knowledge learned in Day 7 to find out the volume of prisms, pyramids, cylinders, cones, and spheres in real life problems.
Prism
area of base × height


Cylinder
πr2h where r = radius, h = height
Cone
1/3
where r = radius, h = height
Pyramid
pyramid
  or area of base x height
                     3
Sphere
sphere
where r = radius


QUIZ


1) Application A contractor needs to build a ramp, as shown at right, from the street to the front of a garage door. How many cubic yards of fill will she need?  


2) If an average rectangular block of limestone used to build the Great Pyramid of Khufu at Giza is approximately 2.5 feet by 3 feet by 4 feet, and limestone weighs approximately 170 pounds per cubic foot, what is the weight of one of the nearly 2,300,000 limestone blocks used to build the pyramid?


3) AQUARIUM TANK The Caribbean Coral Reef Tank at the New England Aquarium is a cylindrical tank that is 23 feet deep and 40 feet in diameter, as shown.
 How many gallons of water are needed to fill the tank? (One gallon of water equals 0.1337 cubic foot.)


Answers









TIPS FROM THE "PROS"







Thursday, May 23, 2013

Unit 2: Day 7

Today, students will learn about the Volume Formulas of Prisms, pyramids, cylinders, cones, and spheres.
Prism
area of base × height
Cylinder
πr2h where r = radius, h = height
Cone
1/3
where r = radius, h = height
Pyramid
pyramid
  or area of base x height
                     3
Sphere
sphere
where r = radius


also the answer is always presented in ans3



QUIZ

1) Find the volume of the prism                  

2) Find the volume of the cone.

3) The radius of a sphere is 6.7 cm.  Find the volume.









ANSWERS



1)  Prism= area of base x height    SO our base is (13 x 3) x (5)...Therefore V=195cm3
2) Cone= pie r2 h=  r=16  h= 34   so (3.14 x (16)2 x 34) divided by 3.   V=9110.2 cm 3
                     3
3) Sphere= 1259.83cm 3
















Unit 2:Day 6

Today, students will learn ALL about Special Right Triangles! Isn't that awesome!

There is 2 types of special right triangles=

45–45–90 triangle

45-45-90 rt triangle







30–60–90 triangle

30-60-90 rt triangle

When solving for problems involving special right triangles, you need to have a calculator and always remember SOHCAHTOA or the pythagorean theorem. 
As SOH= Sin-opposite and Hypotenuse
CAH= Cos-Adjacent and Hypotenuse
TOA= Tan-Opposite +Adjacent











Quiz

ONLY DO 1-3








ANSWERS


1) So Here we are trying to find missing angles. We have and adjacent and an hypotenuse side. So we use COS-1=12/13. So 12 divided by 13 by COS-1.    22.6

2) Here we have an opposite and an adjacent side. So we use Tan-1=4/13
so 4 divided by 13 by TAN-1    17.1

3) Here we have a Hypotenuse and an adjacent angle, so we are using COS. So COS-1=6/9. 6 divided by 9 times COS-1 is....... 48.2










Wednesday, May 22, 2013

Unit 2: Day 5

Today students will learn how to add and subtract radicals.




208+52
First, try to factor any perfect squares out of the radicals.
=208+52
=1613+413
Separate the radicals and simplify.
=1613+413
=413+213
Finally, simplify by combining the terms.
=(4+2)13=613

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

Since the radicals are not the same, and both are in their simplest form, there is no way to combine these values.  The answer is the same as the problem.
Warning:  If the radicals in your problem are different, be sure to check to see if the radicals can be simplified.  Often times, when the radicals are simplified, they become the same radical and can then be added or subtracted.  Always simplify, if possible, before deciding upon your answer.


Quiz

1) Simplify 2√3+3√3


2) 


Simplify:  3√4+2√4

3) 


Simplify:  2√3+3√5





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__________________________________Answer key___________________________






1) 

    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:

    2√3+3√3=(2+3) √3
                 =5√3   
    2) 

  • Simplify:  3√4+2√4
    I have three copies of the radical, plus another two copies,but I can simplify those radicals right down to whole numbers:
      3√4+2√4=3x2+2x2=6+4=10
3) 

  • Simplify:  2√3+3√5
  • These two terms have "unlike" radical parts, and I can't take anything out of either radical. Then I can't simplify the expression 2√3+3√5 any further and my answer has to be:
      2√3+3√5(expression is already fully simplified)