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Pepsi [2]
3 years ago
12

Chris pays a fee if her balance falls below $10 on the statement data. Prior to the statement date, her balance was -$3.46. Then

Chris made a deposit, d, in ample time, so she did not have to pay a fee. Write and solve an inequality to represent this situation. P
Mathematics
1 answer:
alisha [4.7K]3 years ago
3 0

Answer:

The solution of the inequality required for the situation is d  ≥ $13.46.

Step-by-step explanation:

i) the minimum deposit is $10.

ii) the balance before the statement is $-3.46

iii) a deposit d is made so that the fee did not have to be made

iv) therefore d + (-3.46)  ≥  10

                    d - 3.46 ≥ 10

    therefore d  ≥ 10 + 3.46

 therefore d  ≥ $13.46

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F(x) =3x; g(x) = 4x^2 +1
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Which equation is y = –6x^2 + 3x + 2 rewritten in vertex form? y = negative 6 (x minus 1) squared + 8 y = negative 6 (x + one-fo
mart [117]

Answer:

y  = -6(x - \frac{1}{2})^2 -\frac{7}{2}

Step-by-step explanation:

Given:

y = -6x^2 + 3x + 2

Required

Rewrite in vertex form

The vertex form of an equation is in form of: y = a(x - h)^2+ k

Solving: y = -6x^2 + 3x + 2

Subtract 2 from both sides

y - 2 = -6x^2 + 3x + 2 - 2

y - 2 = -6x^2 + 3x

Factorize expression on the right hand side by dividing through by the coefficient of x²

y - 2 = -6(x^2 + \frac{3x}{-6})

y - 2 = -6(x^2 - \frac{3x}{6})

y - 2 = -6(x^2 - \frac{x}{2})

Get a perfect square of coefficient of x; then add to both sides

------------------------------------------------------------------------------------

<em>Rough work</em>

The coefficient of x is \frac{-1}{2}

It's square is (\frac{-1}{2})^2 = \frac{1}{4}

Adding inside the bracket of -6(x^2 - \frac{x}{2}) to give: -6(x^2 - \frac{x}{2} + \frac{1}{4})

To balance the equation, the same expression must be added to the other side of the equation;

Equivalent expression is: -6(\frac{1}{4})

------------------------------------------------------------------------------------

The expression becomes

y - 2 -6(\frac{1}{4})= -6(x^2 - \frac{x}{2} + \frac{1}{4})

y - 2 -\frac{6}{4}= -6(x^2 - \frac{x}{2} + \frac{1}{4})

y - 2 -\frac{3}{2}= -6(x^2 - \frac{x}{2} + \frac{1}{4})

Factorize the expression on the right hand side

y - 2 -\frac{3}{2}= -6(x - \frac{1}{2})^2

y - (2 +\frac{3}{2})= -6(x - \frac{1}{2})^2

y - (\frac{4+3}{2})= -6(x - \frac{1}{2})^2

y - (\frac{7}{2})= -6(x - \frac{1}{2})^2

y  +\frac{7}{2} = -6(x - \frac{1}{2})^2

Make y the subject of formula

y  = -6(x - \frac{1}{2})^2 -\frac{7}{2}

<em>Solved</em>

7 0
3 years ago
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