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2 years ago

6

You are traveling to Japan and need Japanese Yen (JPY). How much JPY could you get for $100 USD if the exchange rate is USD/JPY

0.9333 ?
the options are
¥107.15
¥93.33
¥99.07
¥100.93

2 answers:

Andrew [12]2 years ago

7 0

Answer:

In $100 USD we will get ¥107.15

A is correct.

Step-by-step explanation:

You are traveling to Japan and need Japanese Yen (JPY)

We need to convert $100 USD to JPY

If conversion rate is $1USD = 0.9333JPY

The value of $1 USD is 0.9333 JPY

Let in $100 USD get x JPY

So, The ratio must be same.

$\dfrac{100}{x}=0.9333$

$x=\dfrac{100}{0.9333}$

$x=107.15$

Hence, In $100 USD we will get ¥107.15

Vanyuwa [196]2 years ago

3 0

Answer:

107.15

Step-by-step explanation:

100/.9333=107.15

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Consider the parabola given by the equation: f(x) = 4x² - 6x - 8 Find the following for this parabola: A) The vertex: Preview B)

jeyben [28]

Answer:

The vertex: $(\frac{3}{4},-\frac{41}{4} )$

The vertical intercept is: $y=-8$

The coordinates of the two intercepts of the parabola are $(\frac{3+\sqrt{41} }{4} , 0)$ and $(\frac{3-\sqrt{41} }{4} , 0)$

Step-by-step explanation:

To find the vertex of the parabola $4x^2-6x-8$ you need to:

1. Find the coefficients <em>a</em>, <em>b</em>, and <em>c </em>of the parabola equation

<em> $a=4, b=-6, \:and \:c=-8$ </em>

2. You can apply this formula to find x-coordinate of the vertex

$x=-\frac{b}{2a}$ , so

$x=-\frac{-6}{2\cdot 4}\\x=\frac{3}{4}$

3. To find the y-coordinate of the vertex you use the parabola equation and x-coordinate of the vertex ( $f(-\frac{b}{2a})=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c$ )

$f(-\frac{b}{2a})=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c\\f(\frac{3}{4})=4\cdot (\frac{3}{4})^2-6\cdot (\frac{3}{4})-8\\y=\frac{-41}{4}$

To find the vertical intercept you need to evaluate x = 0 into the parabola equation

$f(x)=4x^2-6x-8\\f(0)=4(0)^2-6\cdot 0-0\\f(0)=-8$

To find the coordinates of the two intercepts of the parabola you need to solve the parabola by completing the square

$\mathrm{Add\:}8\mathrm{\:to\:both\:sides}$

$x^2-6x-8+8=0+8$

$\mathrm{Simplify}$

$4x^2-6x=8$

$\mathrm{Divide\:both\:sides\:by\:}4$

$\frac{4x^2-6x}{4}=\frac{8}{4}\\x^2-\frac{3x}{2}=2$

$\mathrm{Write\:equation\:in\:the\:form:\:\:}x^2+2ax+a^2=\left(x+a\right)^2$

$x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=2+\left(-\frac{3}{4}\right)^2\\x^2-\frac{3x}{2}+\left(-\frac{3}{4}\right)^2=\frac{41}{16}$

$\left(x-\frac{3}{4}\right)^2=\frac{41}{16}$

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PtichkaEL [24]

Answer:

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Step-by-step explanation:

To preface, your figure is going to be a line segment, with $O$ as your midpoint, in between points $P$ & $M.$

With that being said:

$PO+OM=PM$

Identify your values:

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Substitute the values into the first equation:

$7y+12+3y-8=110$

Combine like terms:

$10y+4=110$

Subtract $4$ from both sides of the equation:

$10y=106$

Divide by the coefficient of $y$ , which is $10$ :

$y=10.6$

Substitute $10.6$ for $y$ in segments $PO$ & $OM$ :

$PO=7(10.6)+12$

$OM=3(10.6)-8$

$PO=74.2+12$

$OM=31.8-8$

Solve:

$PO=86.2$

$OM=23.8$

Check your answers by substituting:

$PO+OM=PM$

$86.2+23.8=110$

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