I think the answer would be 1/2 but i’m not sure if that’s what it’s asking
Answer:
$966
Step-by-step explanation:
6/7 = 828/x
Write a proportion.
6x = 7(828) Use cross products.
6x = 5796 Multiply.
x = 966 Divide both sides by 6.
Paul's investment was $966.
That's not correct. The terms 2a and 3b are not like terms, so we cannot combine them to get 5ab. We simply leave it as 2a+3b.
If you had 2a+3a, then it would simplify to 5a
Similarly, 2b+3b = 5b
Or you could have 2ab+3ab = 5ab
The key is that the variable portions must match up to be able to add them.
Given:
The given quadratic polynomial is :
![x^2-x-12](https://tex.z-dn.net/?f=x%5E2-x-12)
To find:
The quadratic polynomial whose zeroes are negatives of the zeroes of the given polynomial.
Solution:
We have,
![x^2-x-12](https://tex.z-dn.net/?f=x%5E2-x-12)
Equate the polynomial with 0 to find the zeroes.
![x^2-x-12=0](https://tex.z-dn.net/?f=x%5E2-x-12%3D0)
Splitting the middle term, we get
![x^2-4x+3x-12=0](https://tex.z-dn.net/?f=x%5E2-4x%2B3x-12%3D0)
![x(x-4)+3(x-4)=0](https://tex.z-dn.net/?f=x%28x-4%29%2B3%28x-4%29%3D0)
![(x+3)(x-4)=0](https://tex.z-dn.net/?f=%28x%2B3%29%28x-4%29%3D0)
![x=-3,4](https://tex.z-dn.net/?f=x%3D-3%2C4)
The zeroes of the given polynomial are -3 and 4.
The zeroes of a quadratic polynomial are negatives of the zeroes of the given polynomial. So, the zeroes of the required polynomial are 3 and -4.
A quadratic polynomial is defined as:
![x^2-(\text{Sum of zeroes})x+\text{Product of zeroes}](https://tex.z-dn.net/?f=x%5E2-%28%5Ctext%7BSum%20of%20zeroes%7D%29x%2B%5Ctext%7BProduct%20of%20zeroes%7D)
![x^2-(3+(-4))x+(3)(-4)](https://tex.z-dn.net/?f=x%5E2-%283%2B%28-4%29%29x%2B%283%29%28-4%29)
![x^2-(-1)x+(-12)](https://tex.z-dn.net/?f=x%5E2-%28-1%29x%2B%28-12%29)
![x^2+x-12](https://tex.z-dn.net/?f=x%5E2%2Bx-12)
Therefore, the required polynomial is
.
Answer:
B) 116°
Step-by-step explanation:
y=180-54-64=62
x=180-62-64=54
62+54=116