As you may already be familiar, these functions f(x) and g(x) are piecewise. They consist of multiple functions with different domains.
1. For #1, the given input is f(0). Since 0≤1, you should use the first equation to solve. f(0)=3(0)-1 ➞ f(0)=-1
2. Continue to evaluate the given input for the domains given. 1≤1, therefore f(1)=3(1)-1➞f(1)=2
3. 5>1, therefore f(5)=1-2(5)➞f(5)=-9
4. -4≤1; f(-4)=3(-4)-1➞f(-4)=-13
5. -3<0<1; g(0)=2
6. -3≤-3; g(-3)=3(-3)-1➞g(-3)=-10
7. 1≥1; g(1)=-3(1)➞g(1)=-3
8. 3≥1; g(3)=-3(3)➞g(3)=-9
9. -5≤-3; g(-5)=3(-5)-1➞g(-5)=-16
Hope this helps! Good luck!
Answer:
6+n/8=2
Step-by-step explanation:
* More Than = (+) or in some cases (>)
* Quotient of a (#) = Any letter that represents a number, so in this case "n", because we do not know that number yet. We divide this with "8" because it says "and 8".
* Equal = <---
Answer:
We need to find which expressions are equivalent to
,
or neither.
: We extract the greatest common factor which is 6. Remember, when we extract a GCM, we divide each term by it.
![6c+12=+(c+2)](https://tex.z-dn.net/?f=6c%2B12%3D%2B%28c%2B2%29)
Therefore, this expression is equivalent to neither of the given expressions.
: We just need to apply the distributive property.
![2(3c+3)=6c+6](https://tex.z-dn.net/?f=2%283c%2B3%29%3D6c%2B6)
Therefore, this expression is equivalen to
.
We use the same process to the other expressions.
![(6c)+(6\times 6)=6c+36=6(c+6)](https://tex.z-dn.net/?f=%286c%29%2B%286%5Ctimes%206%29%3D6c%2B36%3D6%28c%2B6%29)
![3c+6+3c=6c+6](https://tex.z-dn.net/?f=3c%2B6%2B3c%3D6c%2B6)
![6c+36=6(c+6)](https://tex.z-dn.net/?f=6c%2B36%3D6%28c%2B6%29)
, equivalent to neither.
Bring it to the form ax + by = c, where a is positive, and there are no fractions in the equation.
Here, we need to add 2/5x to both sides:
2/5x + y = 0
Then multiply everything by 5 to get rid of the fraction
2x + 5y = 0 <==