I can't really measure the angles. I can tell you what they look like they'd be. There are three reasons why not.
1. The paper is slightly bulged where your hand is
2. The circle is slightly distorted by the camera. This is not that critical.
3. It is very hard to measure on a computer screen.
There is nothing you could do to make it any better. In fact, given what you had to do, this is a pretty good representation.
The Three angles -- Question 22
Using the crudest tools you could imagine, I measured the angle where you have written two 90s between the arms as 132°. That angle opens towards the bottom of the page.
The angle that you have called 95 degrees is actually pretty close. I think you read the upper set of numbers on the protractor when you should have been reading the lower set. I make it 89, but I'll bet it is intended to be 90 degrees.
The third angle on your right is the same as the first one. It comes in at about 132° using my tools again.
Question 23
I can help you with this. When you are asked to make an equation, you have to use an equal sign somewhere.
The sum of the three angles should be 360° I'm going to create an error term because I'm almost sure what I measured won't make 360. All circles when you make angles from any point inside them should make angles that add up to 360° when measured with a protractor if the rays of the angles all start from the same point. [If you don't know what a ray is, call it "the arms of the angles"].
So let's create the equation.
Angle1 + angle2 + angle3 + E = 360°
Angle1 = 89°
Angle2 = 132°
Angle3 = 132°
E is the error that represents the amount away from 360. Your teacher doesn't expect you to get this or to set it into your equation. The main thing you were supposed to do is add up the angles as you tried to do and state what your total was. This is what was expected.
Total = angle1 + angle2 + angle3
Total = 89 + 132 + 132 = 353. My error is 7° too little. So in my equation E = 7°
What to do
Somebody had to mark this with that green felt. You have a teacher. Go to the teacher and ask to be shown how to read the protractor if that person knows. You just need a bit of help. If the teacher cannot tell you, go to someone in your class who knows about that sort of thing and ask them. You're pretty close to getting it.
Answer:
X12/45 =32
Step-by-step explanation:
Answer:
(4/3, 7/3)
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality
<u>Algebra I</u>
- Terms/Coefficients
- Coordinates (x, y)
- Solving systems of equations of using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
7x - y = 7
x + 2y = 6
<u>Step 2: Rewrite Systems</u>
Equation: x + 2y = 6
- [Subtraction Property of Equality] Subtract 2y on both sides: x = 6 - 2y
<u>Step 3: Redefine Systems</u>
7x - y = 7
x = 6 - 2y
<u>Step 4: Solve for </u><em><u>y</u></em>
<em>Substitution</em>
- Substitute in <em>x</em>: 7(6 - 2y) - y = 7
- Distribute 7: 42 - 14y - y = 7
- Combine like terms: 42 - 15y = 7
- [Subtraction Property of Equality] Subtract 42 on both sides: -15y = -35
- [Division Property of Equality] Divide -15 on both sides: y = 7/3
<u>Step 5: Solve for </u><em><u>x</u></em>
- Define original equation: x + 2y = 6
- Substitute in <em>y</em>: x + 2(7/3) = 6
- Multiply: x + 14/3 = 6
- [Subtraction Property of Equality] Subtract 14/3 on both sides: x = 4/3
Answer:
6 swimmers in the first heat can be arranged in 1716 different ways.
Step-by-step explanation:
A swim meet has 13 contestants signed up. To calculate the arrangement of first 6 swimmers in first heat we will use combinations because order doesn't matter.
So to select 6 swimmers out of 13 contestants number of different ways
= ![^{13}C_{6}](https://tex.z-dn.net/?f=%5E%7B13%7DC_%7B6%7D)
= ![\frac{13!}{(6!)(13-6)!}](https://tex.z-dn.net/?f=%5Cfrac%7B13%21%7D%7B%286%21%29%2813-6%29%21%7D)
= ![\frac{13\times 12\times 11\times 10\times 9\times 8\times 7!}{6!\times 7!}](https://tex.z-dn.net/?f=%5Cfrac%7B13%5Ctimes%2012%5Ctimes%2011%5Ctimes%2010%5Ctimes%209%5Ctimes%208%5Ctimes%207%21%7D%7B6%21%5Ctimes%207%21%7D)
= ![\frac{13\times 12\times 11\times 10\times 9\times 8}{6\times 5\times 4\times 3\times 2\times 1}](https://tex.z-dn.net/?f=%5Cfrac%7B13%5Ctimes%2012%5Ctimes%2011%5Ctimes%2010%5Ctimes%209%5Ctimes%208%7D%7B6%5Ctimes%205%5Ctimes%204%5Ctimes%203%5Ctimes%202%5Ctimes%201%7D)
= ![\frac{1235520}{720}](https://tex.z-dn.net/?f=%5Cfrac%7B1235520%7D%7B720%7D)
= 1716
Therefore, 6 swimmers in the first heat can be arranged in 1716 different ways.