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Tems11 [23]
3 years ago
7

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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Is 1.7x^2 + 3.5 a cubic polynomial?
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Answer:

No, it is not.

Step-by-step explanation:

This is because if you look at the expression, there is no term that is raised to the cube root. As a result, this isn't a cubic polynomial. However, this would be a quadratic polynomial cause the highest raised exponent is 2, which is the number for quadratic.

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3 years ago
How do i get 28 and 13 gallons to 17 gallons i really need help with it as soon as i can
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8 0
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


7 0
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
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