Answer:
Step-by-step explanation:
You need to use synthetic division to do all of these. The thing to remember with these is that when you start off with a certain degree polyomial, what you get on the bottom line after the division is called the depressed polynomial (NOT because it has to math all summer!) because it is a degree lesser than what you started.
a. 3I 1 3 -34 48
I'm going to do this one in its entirety so you get the idea of how to do it, then you'll be able to do it on your own.
First step is to bring down the first number after the bold line, 1.
3I 1 3 -34 48
_____________
1
then multiply it by the 3 and put it up under the 3. Add those together:
3I 1 3 -34 48
3
----------------------------
1 6
Now I'm going to multiply the 6 by the 3 after the bold line and add:
3I 1 3 -34 48
3 18
_________________
1 6 -16
Same process, I'm going to multiply the -16 by the 3 after the bold line and add:
3I 1 3 -34 48
3 18 -48
___________________
1 6 -16 0
That last zero tells me that x-3 is a factor of that polynomial, AND that the depressed polynomial is one degree lesser and those numbers there under that line represent the leading coefficients of the depressed polynomial:
![x^2+6x-16=0](https://tex.z-dn.net/?f=x%5E2%2B6x-16%3D0)
Factoring that depressed polynomial will give you the remaining zeros. Because this was originally a third degree polynomial, there are 3 zeros as solutions. Factoring that depressed polynomial gives you the remaining zeros of x = -8 and x = 2
I am assuming that since you are doing synthetic division that you have already learned the quadratic formula. You could use that or just "regular" factoring would do the trick on all of them.
Do the remaining problems like that one; all of them come out to a 0 as the last "number" under the line.
You got this!
Answer:
![L = 30\\R = 21\\M = 42](https://tex.z-dn.net/?f=L%20%3D%2030%5C%5CR%20%3D%2021%5C%5CM%20%3D%2042)
Step-by-step explanation:
Let's assign each person a variable using the first letter of his/her name.
Then we get the following equations based on the information given:
![L + M + R = 93\\L - 9 = R\\M = 2R](https://tex.z-dn.net/?f=L%20%2B%20M%20%2B%20R%20%3D%2093%5C%5CL%20-%209%20%3D%20R%5C%5CM%20%3D%202R)
Then we can use substitution:
![L + M + R = 93\\L + (2R) + R = 93\\L + 3R = 93](https://tex.z-dn.net/?f=L%20%2B%20M%20%2B%20R%20%3D%2093%5C%5CL%20%2B%20%282R%29%20%2B%20R%20%3D%2093%5C%5CL%20%2B%203R%20%3D%2093)
Rearrange the second equation:
![L - R = 9](https://tex.z-dn.net/?f=L%20-%20R%20%3D%209)
Multiple the equation by 3:
![3L - 3R = 27](https://tex.z-dn.net/?f=3L%20-%203R%20%3D%2027)
Combine the equations to solve for a single variable (L):
![L + 3L = 4L\\3R - 3R = 0\\93 + 27 = 120](https://tex.z-dn.net/?f=L%20%2B%203L%20%3D%204L%5C%5C3R%20-%203R%20%3D%200%5C%5C93%20%2B%2027%20%3D%20120)
Then we have a new equation:
![4L = 120\\L = 30](https://tex.z-dn.net/?f=4L%20%3D%20120%5C%5CL%20%3D%2030)
Use substitution to get the rest:
![L - 9 = R\\(30) - 9 = R\\R = 21\\\\M = 2R\\M = 2(21)\\M = 42](https://tex.z-dn.net/?f=L%20-%209%20%3D%20R%5C%5C%2830%29%20-%209%20%3D%20R%5C%5CR%20%3D%2021%5C%5C%5C%5CM%20%3D%202R%5C%5CM%20%3D%202%2821%29%5C%5CM%20%3D%2042)
The area is 144 units.
area is length × width
squares have an equal lenght and width, and 4 sides. if a square's perimeter is 48, we need to divide that by 4 to get the size of each side.
48 ÷ 4 = 12
each side is 12 units. to get the area, we need to mutliple length × width, in this case 12 × 12, which equals 144 units.
Answer:2
Step-by-step explanation:
Answer:
y+48=90
y= 90-48
y= 42
Step-by-step explanation:
ok
by using the property
sum of two opposite interior angles of triangle is equal to exterior angle