Here's a method that you can utilize
If the ratio of the triangle's angle is 2:3:4, that means that the there is an angle with the value 2y, 3y, and 4y. Of course, by the triangle sum theorem, all the angles in a triangle must equal 180 degrees. Now, let's solve for y:
\(2y+3y+4y=180\) | Simplify the left hand side of the equation. 2y+3y+4y=9y. |
\(9y=180\) | Divide by 9 on both sides of the equation. |
\(y=20\) | |
To find the measure of the largest angle, substitute the value for x and plug it into the largest angle, which is 4x since 4 multiplied by a natural number is greater than one multiplied by three:
\(4(20)\) | |
\(y=4*20=80^{\circ}\) | |
2)
This problem is very similar to the one above. This time, however, we must know that the sum of all the exterior angles in any polygon is equal to 360 degrees. We'll use y again and solve just like before:
\(2y+3y+4y=360\) | Combine like terms on the left hand side of the equation. |
\(9y=360\) | Divide by 9 on both sides. |
\(y=40\) | |
Ok, we want the smallest exterior angle to get the largest interior angle:
\(180-(2*40)\) | Evaluate this |
\(180-80\) | |
\(100\) | |
Therefore, the largest interior angle is 100 degrees.
In order to do this problem, you must understand not slope-intercept form but point-slope form.
Point-slope form is the following equation:
\(y-k=m(x-h)\)
m = slope of the line
(h,k) is a point (any point will do) on the line.
We already have all of this information given to us already, so this problem is relatively easy. m = -2 and (h,k) is (4,7). Let's plug this in:
\(y-k=m(x-h)\) | Plug in m as -2, h as 4, and k as 7. |
\(y-7=-2(x-4)\) | You could say that you are done because you have written an equation for the conditions, but let's transform this into slope-intercept form by distributing the -2. |
\(y-7=-2x+8\) | Add 7 to both sides. |
\(y=-2x+15\) | |
To figure this problem out, let's first take a look at all the natural numbers up to 20.
The question asks for the probability for a number to be divisible by 2, 3, or 4. However, 2 is a factor of 4, so any number divisible by 4 is already divisible by 4, so there is no need to ever check for divisibility by 2. Also, if you prove a number is divisible by 2, you do not need to check if a number is divisible by 3.Let's take a look:
Natural Numbers from 1-20 | Divisible by 2? | Divisible by 3? | |||||||
1 | |||||||||
2 | ✔ | ||||||||
3 | ✔ | ||||||||
4 | ✔ | ||||||||
5 | |||||||||
6 | ✔ | ✔ | |||||||
7 | |||||||||
8 | ✔ | ||||||||
9 | ✔ | ||||||||
10 | ✔ | ||||||||
11 | |||||||||
12 | ✔ | ✔ | |||||||
13 | |||||||||
14 | ✔ | ||||||||
15 | ✔ | ||||||||
16 | ✔ | ||||||||
17 | |||||||||
18 | ✔ | ✔ | |||||||
19 | |||||||||
20 | ✔ | ||||||||
Now, let's count how many numbers are divisible by either 2 or 3. There are 13 numbers that divisible by either 2 or 3.
Therefore, there is a \(\frac{13}{20}\) chance that a number selected at random from the natural numbers 1 to 20 is divisible by either 2 or 3.