6+x=c, I’m confused, any help ?

Mathematics

1 answer:

Mashcka [7]3 years ago

3 0

<h3>Explanation</h3>

Solve for x-term

Therefore, we have to make x-term as the subject of equation. Currently c-term is the subject of equation.

$6 + x = c \\ x = c - 6$

Isolate x and we get the answer.

<h3>Answer</h3>

$\large \boxed{x = c - 6}$

You might be interested in

What is the product of (3y^-4)(2y^-4)?

borishaifa [10]

(x^a)(x^b)=x^(a+b)

(ab)(cd)=(a)(b)(c)(d)

x^-m=1/(x^m)

(3y^-4)(2y^-4)=
(3)(y^-4)(2)(y^-4)=
(6)(y^-8)=
6/(y^8)

4 0

3 years ago

Read 2 more answers

The area of the rectangle shown below is 6x2+11x-7 square units. the length is 2x-1 what is the width

Hitman42 [59]

Answer:

Width is 3x+7 units

Step-by-step explanation:

In the attached file

5 0

3 years ago

Read 2 more answers

What are the exact solutions of x2 − 5x − 1 = 0 using x equals negative b plus or minus the square root of the quantity b square

geniusboy [140]

Answer:

$x=2.5+\frac{\sqrt{29}}{2}\\\\x=2.5-\frac{\sqrt{29}}{2}$

Step-by-step explanation:

So the question here is asking you to use the quadratic formula which is expressed as: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\$

A quadratic can generally be expressed as: $y=ax^2+bx+c$

So using the equation you gave: $0=x^2-5x-1$

We can identify the following values: a=1, b=-5, c=-1

Btw the equation explicitly write "1" as the coefficient of x, but since it's not provided it's implied that it's 1.

So plugging in the known values, we get the following equation:

$x=\frac{5\pm\sqrt{(-5)^2-4(1)(-1)}}{2(1)}\\\\x=\frac{5\pm\sqrt{25+4}}{2}\\\\x=\frac{5\pm\sqrt{29}}{2}\\\\x=2.5+\frac{\sqrt{29}}{2}\\\\x=2.5-\frac{\sqrt{29}}{2}$

The last step just consists of taking the + and - solution, and since it asks for exact solutions you leave the 29 under the radical, and you don't approximate. There is no further simplification that can be done here.

5 0

1 year ago

What is the length of the cylinder's height? Please help

raketka [301]

Answer: