The answer is 1.
EXPLANATION
The commutative property of multiplication states that two numbers can be multiplied in either order.
Answer:
the question is cutoff. please rewrite the question. please write exponents as 6x^2.
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:
Anticlockwise 60°
Step-by-step explanation:
Let my starting point be 0°
Turning to the right(clockwise)40° = 0° + 40° = 40°
I am now at 40° to the right of my starting point.
Turning to the left(anticlockwise) 70° = 40° - 70° = -30°
I am now 30° to the left of my starting point.
Turning to the right 90° = -30° + 90° = 60°
I am 60° to the right of my starting point.
To go back to the startoing point(0°), I should go to the left(anticlockwise) by 60°
This is a change of -60°
-Chetan K
Please help me get this answer right
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