All categories

3 years ago

If 2x-1/3=-4x then x=

Mathematics

1 answer:

allochka39001 [22]3 years ago

6 0

Answer:

x = 1/18

Step-by-step explanation:

You might be interested in

Which expression shows

Genrish500 [490]

Answer:

8 less than the product of 4 times 7 is....

8-4×7.

7 0

3 years ago

Read 2 more answers

A bag contains 3 red marbles 4 blue marbles and 7 Green marbles if two marbles are drawn out of the bag what is the probability

mr_godi [17]

Answer:

0.368

Step-by-step explanation:

Number of red marbles= 3

Number of blue marbles= 4

Number of green marbles= 7

Total number of marbles= 3+4+7= 14

Let the first ball is taken out to be red

P(R)= $\frac{3}{14}$

than again a second ball is drawn, number of balls left= 14-1=13

Number of red= 2

P(R)= $\frac{2}{13}$

Total probability= $\frac{3}{14}$ + $\frac{2}{13}$

Total probability= 0.368

Hence, the correct answer is 0.368

3 0

3 years ago

Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.

Flauer [41]

Answer:

16°F

Step-by-step explanation:

15 + x = 31

x = 31 -15

x = 16

4 0

2 years ago

Reyna puts 1/4 pound

gogolik [260]

1/4 pound x 5 days= 1 1/4 pounds (1.25)

4 0

2 years ago

Find all roots: x^3 + 7x^2 + 12x = 0 Show all work and check your answer.

Aliun [14]

The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

Solution:

We have been given a cubic polynomial.

$x^{3}+7 x^{2}+12 x=0$

We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

$x^{3}+7 x^{2}+12 x=0$

$x\left(x^{2}+7 x+12\right)=0$ ----- eqn 1

So, from the above eq1 we can find the first root of the polynomial, which will be:

x = 0

Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

$x^{2}+7 x+12=0$

we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Now, a = 1, b = 7 and c = 12

By substituting the values of a,b and c in the quadratic equation we get;

$\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}$

Therefore, the two roots are:

$\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}$

And,

$\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}$

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

4 0

3 years ago