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Natali [406]
3 years ago
6

Solve for x.

Mathematics
2 answers:
denis-greek [22]3 years ago
5 0

Answer:

\frac{3\±\sqrt{3}}{3}

Step-by-step explanation:

Whe have the function 3x^2-6x+2=0

To solve this equation we must factor.

To factor, we need to find the points where the function is equal to zero. So we use the quadratic formula

\frac{-b\±\sqrt{b^2-4ac}}{2a}

Where:

a = 3

b = -6

c = 2

Then

\frac{-(-6) + \sqrt{(-6)^2-4(3)(2)}}{2(3)} = \frac{3+\sqrt{3}}{3}\\\\and\\\\\frac{-(-6) - \sqrt{(-6)^2-4(3)(2)}}{2(3)}= \frac{3-\sqrt{3}}{3}

finally the correct answer is  \frac{3\±\sqrt{3}}{3}

lisov135 [29]3 years ago
3 0

just made this so the other person could get brainiest as a thanks.

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Answer:

1\frac{3}{4}

Step-by-step explanation:

In order to find the answer, you first need to convert the mixed number (2 1/3) into a improper fraction and then multiply.

\frac{3}{4} \times2\frac{1}{3}

2\frac{1}{3} =3\times2=6+1=7=\frac{7}{3}

\frac{3}{4} \times\frac{7}{3}

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Hope this helps.

7 0
2 years ago
What is 30 less than 4 times a number is equal to 14
Genrish500 [490]
The answer is x=11
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You add 30 to each side=4x=44
Then you divide 44 by 4 which=11
4 0
3 years ago
The volunteers at a high school football team’s concession stand are trying to decide on the price of the hot dogs they are sell
olga55 [171]

Answer:

x is the number of $1 increase in the price.

If there is no increase, then the total money earned is

2 × 70 = 140

If there is $1 increase, then the total money earned is

(2 + 1) × [70 - 8(1)]

If there is $2 increase, then the total money earned is

(2 + 2) × [70 - 8(2)]

If we continue the pattern, for x times $1 increase, total money earned is

(2 + x)(70 - 8x) =

If we substitute x = 0 in the above equation, we will get

the total money earned = $140.

It means if there is no increase, then the total money earned = 140.

Hence, 140 is the constant term and it represents that there is no increase in price.

Still stuck? Get 1-on-1 help from an expert tutor now.

Step-by-step explanation:

6 0
2 years ago
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