All categories

3 years ago

Variable termes X-12-3X+22

Mathematics

1 answer:

adell [148]3 years ago

7 0

Answer: x + 12 = 3x + -22

12 + x = 3x + -22

Reorder the terms:

12 + x = -22 + 3x

Solving

12 + x = -22 + 3x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-3x' to each side of the equation.

12 + x + -3x = -22 + 3x + -3x

Combine like terms: x + -3x = -2x

12 + -2x = -22 + 3x + -3x

Combine like terms: 3x + -3x = 0

12 + -2x = -22 + 0

12 + -2x = -22

Add '-12' to each side of the equation.

12 + -12 + -2x = -22 + -12

Combine like terms: 12 + -12 = 0

0 + -2x = -22 + -12

-2x = -22 + -12

Combine like terms: -22 + -12 = -34

-2x = -34

Divide each side by '-2'.

x = 17

Simplifying

x = 17

You might be interested in

If i can get 2 prego pasta sauce for $8 how much would it be for 1 prego pasta?

Sphinxa [80]

Answer: $4

Step-by-step explanation:

If you have 2 Prego pasta sauces at $8 and you only want half of that (1), you divide 8 by 2 to get 4. 1 pasta sauce equals $4

3 0

2 years ago

Read 2 more answers

X^6+3x^3-5 <br> write the expression in quadratic form

Art [367]

Step-by-step explanation:

$v = {x}^{3} \\ {v}^{2} + 3v - 5 = 0 \\$

Solving this equation using the quadratic formula, we get two real solutions :

1.1926 or -4.1926

Now we know the values of v , we can calculate x since x is ∛ v

${x}^{6} + 3 {x}^{3} - 5 = 0$

$x = \sqrt[3]{1.1926} = 1.0605 \\ x = \sqrt[3]{ - 4.1926} = - 1.6125$

3 0

3 years ago

RSM wants to send four of its 18 Math Challenge teachers to a conference. How many combinations of four teachers include exactly

tatiyna

Answer:

1120 combinations of four teachers include exactly one of either Mrs. Vera or Mr. Jan.

Step-by-step explanation:

The order in which the teachers are chosen is not important, which means that the combinations formula is used to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

1 from a set of 2(Either Mrs. Vera or Mr. Jan).

3 from a set of 18 - 2 = 16. So

$C_{2,1}C_{16,3} = \frac{2!}{1!1!} \times \frac{16!}{3!13!} = 1120$

1120 combinations of four teachers include exactly one of either Mrs. Vera or Mr. Jan.

4 0

3 years ago

Glven: 2x+3y= 6.

ruslelena [56]

Answer:

The answer is B 0,2

Step-by-step explanation:

I hope that correct

7 0

3 years ago

For the arthritic sequence 5,9,13,17,21,... how many terms would you need to get to 129

Vlad1618 [11]

9514 1404 393

Answer:

Step-by-step explanation:

The first term is 5 and the common difference is 9-5=4, so the n-th term is ...

an = a1 +d(n -1)

an = 5 +4(n -1) = 4n +1

Then the term number for the term 129 is ...

129 = 4n +1 . . . . . put the given value in the formula

128 = 4n . . . . . . . subtract 1

32 = n . . . . . . . . . .divide by 4

The 32nd term is 129.

8 0

3 years ago

Variable termes X-12-3X￼￼+22

Variable termes X-12-3X+22