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3 years ago

A. Write the equation ax + b = c in terms of x.

Mathematics

2 answers:

Oksanka [162]3 years ago

5 0

Answer:

x=c/a-b/a

Step-by-step explanation:

ax+b=c

1) Subtract b from both sides:

ax=c-b

2) Divide both sides by a:

x=c/a-b/a

Likurg_2 [28]3 years ago

4 0

Ax + b = c
ax = c - b (subtraction property of equality)
x = (c-b)/a (división property of equality)

P.S. I would really appreciate it if you gave me brainliest :)

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Answer:

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2 years ago

Is 1.7x^2 + 3.5 a cubic polynomial?

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Answer:

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Step-by-step explanation:

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2 years ago

7v=2v-20 what is the answer

Ierofanga [76]

7v = 2v - 20

First, subtract 2v from both sides. / Your problem should look like: 7v - 2v = -20
Second, simplify 7v - 2v to 5v. / Your problem should look like: 5v = -20
Third, divide both sides by 5. / Your problem should look like: v = $- \frac{20}{5}$
Fourth, simplify $\frac{20}{5}$ to 4. / Your problem should look like: v = -4

Answer: v = -4

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3 years ago

Solving a Two-Step Matrix Equation<br> Solve the equation:

Cloud [144]

Answer:

$\boxed {x_{1} = 3}$

$\boxed {x_{2} = -4}$

Step-by-step explanation:

Solve the following equation:

$\left[\begin{array}{ccc}3&2\\5&5\\\end{array}\right] \left[\begin{array}{ccc}x_{1}\\x_{2}\\\end{array}\right] + \left[\begin{array}{ccc}1\\2\\\end{array}\right] = \left[\begin{array}{ccc}2\\-3\\\end{array}\right]$

-In order to solve a pair of equations by using substitution, you first need to solve one of the equations for one of variables and then you would substitute the result for that variable in the other equation:

-First equation:

$3x_{1} + 2x_{2} + 1 = 2$

-Second equation:

$5x_{1} + 5x_{2} + 2 = -3$

-Choose one of the two following equations, which I choose the first one, then you solve for $x_{1}$ by isolating

$3x_{1} + 2x_{2} + 1 = 2$

-Subtract $1$ to both sides:

$3x_{1} + 2x_{2} + 1 - 1 = 2 - 1$

$3x_{1} + 2x_{2} = 1$

-Subtract $2x_{2}$ to both sides:

$3x_{1} + 2x_{2} - 2x_{2} = -2x_{2} + 1$

$3x_{1} = -2x_{2} + 1$

-Divide both sides by $3$ :

$3x_{1} = -2x_{2} + 1$

$x_{1} = \frac{1}{3} (-2x_{2} + 1)$

-Multiply $-2x_{2} + 1$ by $\frac{1}{3}$ :

$x_{1} = \frac{1}{3} (-2x_{2} + 1)$

$x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}$

-Substitute $-\frac{2x_{2} + 1}{3}$ for $x_{1}$ in the second equation, which is $5x_{1} + 5x_{2} + 2 = -3$ :

$5x_{1} + 5x_{2} + 2 = -3$

$5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3$

Multiply $-\frac{2x_{2} + 1}{3}$ by $5$ :

$5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3$

$-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3$

-Combine like terms:

$-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3$

$\frac{5}{3}x_{2} + \frac{11}{3} = -3$

-Subtract $\frac{11}{3}$ to both sides:

$\frac{5}{3}x_{2} + \frac{11}{3} - \frac{11}{3} = -3 - \frac{11}{3}$

$\frac{5}{3}x_{2} = -\frac{20}{3}$

-Multiply both sides by $\frac{5}{3}$ :

$\frac{\frac{5}{3}x_{2}}{\frac{5}{3}} = \frac{-\frac{20}{3}}{\frac{5}{3}}$

$\boxed {x_{2} = -4}$

-After you have the value of $x_2$ , substitute for $x_{2}$ onto this equation, which is $x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}$ :

$x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}$

$x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}$

-Multiply $-\frac{2}{3}$ and $-4$ :

$x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}$

$x_{1} = \frac{8 + 1}{3}$

-Since both $\frac{1}{3}$ and $\frac{8}{3}$ have the same denominator, then add the numerators together. Also, after you have added both numerators together, reduce the fraction to the lowest term:

$x_{1} = \frac{8 + 1}{3}$

$x_{1} = \frac{9}{3}$

$\boxed {x_{1} = 3}$

5 0

2 years ago