All categories

3 years ago

What are the solutions to the equation (2x – 5)(3x – 1) = 0?

Mathematics

2 answers:

Simora [160]3 years ago

8 0

x = 5/2 or x = 1/3

C is the answer.

bezimeni [28]3 years ago

5 0

Answer:

x = 5/2 or x = 1/3

Step-by-step explanation:

Simplify

( 2 x − 5 )( 3 x − 1 ) .

6 x 2 − 17 x + 5 = 0

Use the quadratic formula to find the solutions.

$\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $\frac{-b-\sqrt{b^2-4ac} }{2a}$

Substitute the values a = 6 , b = − 17 , and c = 5 into the quadratic formula and solve for x .

17 ±sqrt(-17)^2-4*(6*5)

_________________

2*6

Simplify.

x = 17 ± 13 /12

The final answer is the combination of both solutions.

x = 5 /2 , 1 /3

You might be interested in

The total cost after tax to repair Deborah’s computer is represented by 0.09(60h)+60h, where h represents the number of hours it

tia_tia [17]

Answer:

0.09 OR 9% is her tax

Step-by-step explanation:

Deborah has to pay 9% for tax. So is someone worked on her computer for two hours then she pays 9% on $120 worth of work. So her tax amount would be $10.800 and her total amount would be 10.8 +120 = $130.80.

4 0

4 years ago

A bag contains red marbles, white marbles, and blue marbles. Randomly choose two marbles, one at a time, and without replacement

dsp73

Answer:

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

$P(Same) = \frac{67}{210}$

Step-by-step explanation:

Given (Omitted from the question)

$Red = 7$

$White = 9$

$Blue = 5$

Solving (a): $P(First\ White\ and\ Second\ Blue)$

This is calculated using:

$P(First\ White\ and\ Second\ Blue) = P(White) * P(Blue)$

$P(First\ White\ and\ Second\ Blue) = \frac{n(White)}{Total} * \frac{n(Blue)}{Total - 1}$

<em>We used Total - 1 because it is a probability without replacement</em>

So, we have:

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{21 - 1}$

$P(First\ White\ and\ Second\ Blue) = \frac{9}{21} * \frac{5}{20}$

$P(First\ White\ and\ Second\ Blue) = \frac{9*5}{21*20}$

$P(First\ White\ and\ Second\ Blue) = \frac{45}{420}$

$P(First\ White\ and\ Second\ Blue) = \frac{3}{28}$

Solving (b) $P(Same)$

This is calculated as:

$P(Same) = P(First\ Blue\ and Second\ Blue)\$ $or\ P(First\ Red\ and Second\ Red)\ or\ P(First\ White\ and Second\ White)$

$P(Same) = (\frac{n(Blue)}{Total} * \frac{n(Blue)-1}{Total-1})+(\frac{n(Red)}{Total} * \frac{n(Red)-1}{Total-1})+(\frac{n(White)}{Total} * \frac{n(White)-1}{Total-1})$