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4 years ago

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Which number(s) below belong to the solution set of the equation? Check all that apply.x + 12 = 13A.14B.1C.13D.312E.0F.25

1 answer:

Darya [45]4 years ago

4 0

X + 12 = 13
x = 13 - 12
x = 1

The only solution is x = 1
letter B

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A person on tour has dollar 360 for his daily expenses. If he extends his tour for 4 days, he has to cut down his daily expenses

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Answer:

The original duration of the tour = 20 days

Step-by-step explanation:

Solution:

Total expenses for the tour = $360

Let the original tour duration be for $x$ days.

So, for $x$ days the total expense = $360

<em>Thus the daily expense in dollars can be given by</em> = $\frac{360}{x}$

Tour extension and effect on daily expenses.

The tour is extended by 4 days.

<em>Tour duration now</em> = $(x+4)$ days

On extension, his daily expense is cut by $3

<em>New daily expense in dollars </em>= $(\frac{360}{x}-3)$

Total expense in dollars can now be given as: $(x+4)(\frac{360}{x}-3)$

Simplifying by using distribution (FOIL).

$(x.\frac{360}{x})+(x(-3)+(4.\frac{360}{x})+(4(-3))$

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We know total expense remains the same which is = $360.

So, we have the equation as:

$348-3x+\frac{1440}{x}=360$

Multiplying each term with $x$ to remove fractions.

$348x-3x^2+1440=360x$

Subtracting $348x$ both sides

$348x-348x-3x^2+1440=360x-348x$

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Dividing each term with -3.

$\frac{-3x^2}{-3}+\frac{1440}{-3}=\frac{12x}{-3}$

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Adding $4x$ both sides.

$x^2+4x-480=-4x+4x$

$x^2+4x-480=0$

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For a quadratic equation: $ax^2+bx+c=0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Plugging in values from the equation we got.

$x=\frac{-4\pm\sqrt{(4)^2-4(1)(-480)}}{2(1)}$

$x=\frac{-4\pm\sqrt{16+1920}}{2}$

$x=\frac{-4\pm\sqrt{1936}}{2}$

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