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Fudgin [204]
3 years ago
13

Ali, Carrie and Bryan received a sum of money. Bryan's money was 3/5 as much as Ali's money. The ratio of Ali's money to Carrie'

s money was 4:1. Ali had $160 more than Bryan. How much was the sum of money.
Mathematics
1 answer:
monitta3 years ago
4 0

Answer:

The sum of money received by Ali, Carrie and Bryan is $ 740.

Step-by-step explanation:

At first we translate mathematically each sentence:

(i) <em>Ali, Carrie and Bryan received a sum of money. </em>

a - Ali's money.

b - Bryan's money.

c - Carrie's money.

(ii) <em>Bryan's money was </em>\frac{3}{5}<em> of  Ali's money</em>.

b = \frac{3}{5}\cdot a (1)

(iii) <em>The ratio of Ali's money to Carrie's money was 4 : 1</em>.

\frac{a}{c} = 4 (2)

(iv) <em>Ali had $ 160 more than Bryan</em>.

a = b + 160 (3)

After some algebraic handling, we have the following system of linear equations:

3\cdot a - 5 \cdot b = 0 (1b)

a - 4\cdot c = 0 (2b)

a - b = 160 (3b)

The solution of the system is: a = 400, b = 240, c = 100

The sum of money is:

s = a + b + c

s = 740

The sum of money received by Ali, Carrie and Bryan is $ 740.

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Which statement describes the inverse of m(x) = x2 – 17x?
stealth61 [152]

Answer:

The correct option is;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

Step-by-step explanation:

The given information is that m(x) = x² - 17·x

The above equation can be written in the form;

y = x² - 17·x

Therefore;

0 = x² - 17·x - y

From the general solution of a quadratic equation, 0 = a·x² + b·x + c we have;

x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}

By comparison to the equation,0 = x² - 17·x - y, we have;

a = 1, b = -17, and c = -y

Substituting the values of a, b and c into the formula for the general solution of a quadratic equation, we have;

x = \dfrac{-(-17)\pm \sqrt{(-17)^{2}-4\times (1) \times (-y)}}{2\times (1)} = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}

Which can be simplified as follows;

x =  \dfrac{17\pm \sqrt{289+4\cdot y}}{2}= \dfrac{17}{2} \pm \dfrac{1}{2}  \times \sqrt{289+4\cdot y}} = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +\dfrac{4\cdot y}{4} }}

And further simplified as follows;

x = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +y }} = \dfrac{17}{2} \pm \sqrt{y + \dfrac{289}{4} }}

Interchanging x and y in the function of the inverse, m⁻¹(x), we have;

m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

We note that the maximum or minimum point of the function, m(x) = x² - 17·x found by differentiating the function and equating the result to zero, gives;

m'(x) = 2·x - 17 = 0

x = 17/2

Similarly, the second derivative is taken to determine if the given point is a maximum or minimum point as follows;

m''(x) = 2 > 0, therefore, the point is a minimum point on the graph

Therefore, as x increases past the minimum point of 17/2, m⁻¹(x) increases to give;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }} to increase m⁻¹(x) above the minimum.

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2 years ago
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Answer:

<h2>-1</h2>

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5x=3x+20
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-Twix
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