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Grace [21]
3 years ago
14

your grandmother got out of the hospital she was so excited she gave you some money you decided to go to Canada where the exchan

ge rate is $1.06 Canadian dollars to $1 US dollar as of November 2013. You have $10000 to take with you. How many Canadians dollars do you have to spend?
Mathematics
1 answer:
Kobotan [32]3 years ago
6 0

Answer:

You have $10.600 Canadian dollars to spend.

Step-by-step explanation:

This problem can be solved by a rule of three.

In a rule of three problem, the first step is identifying the measures and how they are related, if their relationship is direct of inverse.

When the relationship between the measures is direct, as the value of one measure increases, the value of the other measure is going to increase too.

When the relationship between the measures is inverse, as the value of one measure increases, the value of the other measure will decrease.

In this problem, we have the following measures:

-The number of US dollars

-The number of Canadian dollars.

As the number of US dollars increases, so does the number of Canadian dollars, which means that the relationship between the measures is direct.

The problem states that $1 US dollar is $1.06 Canadian dollars. The problme wants to know how much $10000 is in Canadian dollars. So:

1 - 1.06

10000 - x

x = 10000*1.06

x = $10.600

You have $10.600 Canadian dollars to spend.

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Answer:

The coordinates of the image are

W'(2,-4)

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Step-by-step explanation:

we know that

The translation is 8 units  right and 3 units down

so

The rule of the translation is

(x,y) ----> (x+8,y-3)

Apply the rule of the translation to the coordinates of the pre-image WXYZ to obtain the coordinates of the image W'X'Y'Z'

coordinate W'

W(-6, -1) ----> W'(-6+8,-1-3)

W(-6, -1) ----> W'(2,-4)

 coordinate X'

X(-5, -3) ----> X'(-5+8,-3-3)

X(-5, -3) ----> X'(3,-6)

coordinate Y'

Y(-2, 3) ----> Y'(-2+8,3-3)

Y(-2, 3) ----> Y'(6,0)

coordinate Z'

Z(-3, 1) ----> Z'(-3+8,1-3)

Z(-3, 1) ----> Z'(5,-2)

therefore

The coordinates of the image are

W'(2,-4)

X'(3,-6)

Y'(6,0)

Z'(5,-2)

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The main identity you need is the double angle one for cosine:

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We get

\cos^6x=(\cos^2x)^3=\left(\dfrac{1+\cos2x}2\right)^3=\dfrac{(1+\cos2x)^3}8

Expand the numerator to apply the identity again:

\cos^6x=\dfrac{1+3\cos2x+3\cos^22x+\cos^32x}8

\cos^6x=\dfrac{1+3\cos2x+3\left(\frac{1+\cos2(2x)}2\right)+\cos2x\left(\frac{1+\cos2(2x)}2\right)}8

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