Answer:
a. 2.5
b. 2.333
c. 2.85
Step-by-step explanation:
A) 1. subtract -x on both sides 2. to get 3x-1/2=7
3. Add 1/2 on both sides 4. to get 3x=7.5
5. divide 3 on both sides 6. To get... x=2.5
B)1. Same process. 2. x+2=13/3
3. x=13/3-2 4. x= 2.33(repeating)
C) 1. same process. 2. 2x-x/3=3+7/4
3. 2x-x/3=4.75 4. Isolate -x/3
5.-x/3=4.75-2x 6. multiply both sides by 3
7. -x=14.25-6x 8. 5x=14.25
9. x= 2.85
* remember to distribute the 3 to each variable.
Answer:
A
Step-by-step explanation:
easy way to take a note
is that the sum of the 2 first angles much be greater than the 3rd one
so A
Use the rational roots test. The possible roots are: plus/minus 6,3,2,1
Use synthetic division and you will see that 3 is a root:
3 | 1 -3 -3 11 -6
| 3 0 -9 6
____________
1 0 -3 2 0
Use rational root again, to see that possible roots are: plus/minus 2,1
Try 2:
2 | 1 0 -3 -2
| 2 4 2
_____________
1 2 1 0
The above is x^2+2x+1 which is a perfect square: (x+1)^2
So we have the final factorization: (x-3)(x-2)(x+1)^2
So the roots are: 3, 2, -1
Where -1 is a double zero.
Step-by-step explanation:
According to this trigonometric function, −C gives you the OPPOSITE terms of what they really are, so be EXTREMELY CAREFUL:
![\displaystyle Phase\:[Horisontal]\:Shift → \frac{\frac{π}{3}}{3} = \frac{π}{9} \\ Period → \frac{2}{3}π](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Phase%5C%3A%5BHorisontal%5D%5C%3AShift%20%E2%86%92%20%5Cfrac%7B%5Cfrac%7B%CF%80%7D%7B3%7D%7D%7B3%7D%20%3D%20%5Cfrac%7B%CF%80%7D%7B9%7D%20%5C%5C%20Period%20%E2%86%92%20%5Cfrac%7B2%7D%7B3%7D%CF%80)
Therefore we have our answer.
Extended Information on the trigonometric function
![\displaystyle Vertical\:Shift → D \\ Phase\:[Horisontal]\:Shift → \frac{C}{B} \\ Period → \frac{2}{B}π \\ Amplitude → |A|](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Vertical%5C%3AShift%20%E2%86%92%20D%20%5C%5C%20Phase%5C%3A%5BHorisontal%5D%5C%3AShift%20%E2%86%92%20%5Cfrac%7BC%7D%7BB%7D%20%5C%5C%20Period%20%E2%86%92%20%5Cfrac%7B2%7D%7BB%7D%CF%80%20%5C%5C%20Amplitude%20%E2%86%92%20%7CA%7C)
NOTE: Sometimes, your vertical shift might tell you to shift your graph below or above the <em>midline</em> where the amplitude is.
I am joyous to assist you anytime.
