2 of course 1+1=2 ik what your talking ab
X^2 + 3x - 6
This is because x^2 and 3x have nothing they can combine with, leaving them as their own number. 1 + - 6 will give you -5 as adding to a negative only reduces not adds entirely.
Hope this helps!
The answer is negative four. subtract six from both sides and you get -20=5x then you would divide both sides by 5 and get -4
Answer:
<h2><em>
B. (b+3c)+(b+3c) </em></h2><h2><em>C. </em><em>
2(b)+2(3c)</em></h2>
Step-by-step explanation:
Given this expression 2(b+3c), its equivalent expression is derived by simply opening up the bracket as shown below;
Open the parenthesis by multiplying the constant outside the bracket with all the variables in parenthesis.
= 2(b+3c)
= 2(b)+ 2(3c)
= 2b +2*3*c
= 2b +6c
It can also be written as sum of b+3c in 2 places i.e (b+3c)+(b+3c) because multiplying the function b+3c by 2 means we are to add the function by itself in two places.
<em>Hence the equivalent expression are (b+3c)+(b+3c) and 2(b)+2(3c) or 2b+6c</em>
Answer:
Step-by-step explanation:
Given
Required
Determine the area with coordinates 
The area is represented as:
Where
and
Substitute values for r, a and b in
Expand
By integratin the above, we get:
![Area = \frac{1}{2}*\frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{2}[0,2]](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2A%5Cfrac%7B%28cos%28%5Ctheta%29%20%2B%204%29sin%28%5Ctheta%29%20%2B%203%5Ctheta%7D%7B2%7D%5B0%2C2%5D)
![Area = \frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{4}[0,2]](https://tex.z-dn.net/?f=Area%20%3D%20%5Cfrac%7B%28cos%28%5Ctheta%29%20%2B%204%29sin%28%5Ctheta%29%20%2B%203%5Ctheta%7D%7B4%7D%5B0%2C2%5D)
Substitute 0 and 2 for 
one after the other
Get sin(2) and cos(2) in radians
