We will return to rotation and dilation when classes resume hopefully sooner than later. The subject is harder to teach via this online method. I will introduce two new types of problems to solve that incorporate algebra and geometry this week. Much of what we are working on going forward, is a review for the final. Graduation is in 49 days. Your final and this is an approximation based on last years class is 40 days from today, classroom wise.
First we will review transversal’s. Remember all lines equal 180 degrees. If one part of the line is 50 degrees to find the missing degree, simply subtract 50 from 180, this gives you 130 the missing degree.
Many of the answers are repetitive regarding terms. The term congruent play a major role on page 348. We will review the different types of angles on page 349-352. These are topics we worked on last year as well as this year.
Pages 355 -356 ask you to find the missing angles. Add the given degrees and subtract from 180 degrees.
On pages 357-359 the use the same concept that everything equals 180 degrees however, they add a component of algebra to geometry. I will do the problem on page 357 Your term.
Add like terms (5y+3) + ( 4y+8) 5y+4y= 9y 3+8=11…. 5y+11=146, The 146 represents the extended line attached to the triangle.
Set up into an algebraic equations : 9y+11=146; subtract 11 from both sides : 9y=135;
Divide both sides by 9y. Y=15. Next, take the value of 15 and place it back into the original equations. 5(15)+3= 78 degrees. 4(15)+8=68. Your answers are 78 and 68 degrees.
If you any questions feel free to contact either the we posted last week or through Google Classroom. Take care!
Seventh Grade, we are going to skip some chapters, because they are more involved and really need to be taught in the class room. We will work this week on the area and circumference of a circle. Obviously, circles have no edges. They do however; have something called cords, part of the curved line of a circle that is illustrated by dots on the circle. To find the circumference multiply the diameter (The line that extends from one side of the circle to another), it is illustrated on page 265. The Radius is one half of the diameter! To find the circumference of a circle, multiply the diameter times Pi (3.14). If they give you the radius and ask you to find the circumference you need to double the radius and then multiply that times the radius.
If they are looking to find the area the formula is: radius squared times Pi. If they are looking for the area and the give you the diameter, do the following: take half of the diameter, square that number and then multiply times Pi.
If you need any help feel free to text the way you had last week or through Google classroom. If you want we may conduct a class at a certain time to help you do your work this week and going forward, please let me know.
This week we will start with a review of PEMDAS and then we will begin writing equations. Writing equations is an early introduction to algebra and that is introduced with the problem on page 294 number 11. The problems reads: 17 is the sum of a number ans six. 17 is the end number when you add six and another number. Therefore, 17=x+6. They are looking for you to create an equation and it was created with, 17=x+6. If they asked us to solve the question you would solve like this: 17=x+6-6…..You subtract 6 from the left side to unite it with the whole number 17 by subtracting six from the 17.. The rule is whole numbers with whole numbers and variables with variables, and then, divide the variable into the whole number. In this problem we subtract six, because it was positive. You do the opposite of the sign when you are moving across the equal sign in this case. If the six was negative you would have added six to both sides. Continuing: 17=x+6-6 equals 17-6=x equals 17-6 =11 therefore, 11=x or you might write x=11, it’s the same answer. If we took the answer 11 and added the six as a check, we would see 11+6=17.
On Wednesday, on page 300, they will ask you whether the equation is correct when you add a certain number in place of the variable, for example. 14+x=46; x=32. They are looking for you to add 32 in place of the x and then add 14 to see if that equals 46. 14+32= 46 therefore, this is a solution. Another example: 17y=85; y=5. When a number a a variable attached it means you have to multiply. 17(5) =85, and 85=85 therefore, this is a solution. Sometimes the answer is this is not a solution. 5q=31; q=13. If we multiply 5 times 31 we get 155 and not 31 therefore, this is not a solution.
If you have any questions please contact me and have no worries!
This week we will be reviewing The Order of Operations (PEMDAS). Please remember what the Acronym represents. P is for Parentheses, E is for exponents, M is for Multiplication, D is for Division, A is for Addition, and S is for subtraction. Also, remember that Multiplication and Division are equal. With PEMDAS we work from left to right therefore, if the division aspect of the problems comes first, you would divide. If the Multiplication part comes first you would multiply. The same rule applies to Addition and Subtraction.
If you have any problems, please contact me either the same way you posted your work last week or through Google Classwork. If you would like to communicate a class live we may do that as well. This is a learning process for all concerned. Certainly I would rather be in the classroom at OLGCA!