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Alenkasestr [34]
3 years ago
15

What are the values of a, b, and c in the quadratic equation 0 = 1/2 x2 – 3x – 2?

Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
8 0
In a quadratic equation, the standard equation is represented as follows:

Ax^2 + Bx + C = 0

where A, B and C are constants and A should not have a zero value.

The values of A, B, and C would be as follows:

A = 1/2
B = -3
C = -2

Hope this answers the question. Have a nice day.
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X/4 - x/6 = 1<br><br> please help
Archy [21]
To solve this multiply through by the LCM of 4 and 6, which is 12:-

12 * x/4 - 12 * x/6 = 1*12

3x - 2x = 12

x = 12  answer
4 0
3 years ago
I want the answers please for all questions
Sindrei [870]
7. A and B = 4
8. A and C = 8
9. B and D = 5
10. C and G = 10
11. D and F = 6
12. E and F = 5
13. E and B = 6
14. E and A = 10
15. E and G = 8
16. F and G = 13

Thats all I know,  Good Luck <span> </span>



8 0
3 years ago
How do you solve 2(x+7)+x=20
alexira [117]
2(x+7) + x=20
(2)(x) + (2)(7) + x=20  Distribute 
2x+14 + x =20
(2x+x) + (14) =20  Combine Like Terms 
3x+14=20 
    - 14  -14             Subtract 14 from both sides 
3x = 6 
3x/3 6/3                  Divide Both Sides by 3 
 
x = 2 


Let me know if you still don't understand 

4 0
3 years ago
If the outliers are not included, what is the mean of the data? 76,79,80,82,50,78,83,79,81,82.
natta225 [31]
50 is the only outlier so take 50 out of the set then add the other 9 data points and divide by 9.<span>
76 + 79 + 80 + 82 + 78 + 83 + 79 + 81 + 82 = 720 / 9 = 80</span>
4 0
3 years ago
Solving Exponential and Logarithmic Equations In Exercise, solve for x.<br> In 2x - In(3x - 1) = 0
LenKa [72]

Answer:

\frac{-1}{3x^2-x}

Step-by-step explanation:

  1. If f(x) is in th form of f(x)=g(x)-h(x) then f'(x)=g'(x) - h'(x)
  2. When f(x)=z(g(x)) then f'(x)= z'(g(x))g'(x) (called as chain rule)

<u>using these information</u>:

g(x)=ln2x then g'(x)=\frac{(2x)'}{2x} =\frac{2}{2x}=\frac{1}{x}

h(x)=In(3x - 1) then h'(x)=\frac{(3x-1)'}{3x-1} =\frac{3}{3x-1}f'(x)=g'(x) - h'(x) =[tex]\frac{1}{x} - \frac{3}{3x-1} =\frac{-1}{3x^2-x}

7 0
4 years ago
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