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2 years ago

Suppose that the relation G is defined as follows.

Mathematics

1 answer:

Serga [27]2 years ago

6 0

G:{3,4,6}->{0,9}

The pairs represent the input (first number in each pair) and the result (second nu.ber in each pair) for the relation G. for example G(3)=9.

The domain is the set of values that the relation can act upon. The range is the set of the values the results can take

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$3^5=243$

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If a^2 + b^2=c^2 then the triangle is a right triangle.

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What is the value of the exponential expression 16 1/2

Oksana_A [137]

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ Anything to the power of a half means you square root it

The square root of 16 is 4

Therefore, your answer is 4

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3 0

3 years ago

Please help I have spent hours trying to do...

Igoryamba

Answer:

914.4 mm
0.0479594 km

Step-by-step explanation:

For units conversion, use a conversion factor ratio with the units you do want in the numerator and the units you don't want in the denominator. The numbers in the numerator and denominator give the ratio a value of 1: the numerator is equal to the denominator--just in different units.

1. For feet to mm, you need a couple of additional conversion factors besides the ones shown:

1 foot = 12 in

1 cm = 10 mm

So, the conversion is ...

$(3\text{ ft})\times\dfrac{12\text{ in}}{1\text{ ft}}\times\dfrac{2.54\text{ cm}}{1\text{ in}}\times\dfrac{10\text{ mm}}{1\text{ cm}}=3\times12\times2.54\times10\text{ mm}\\\\=\boxed{914.4\text{ mm}}$

2. The additional factor you need for this problem is ...

1 mile = 5280 feet

So, the conversion is ...

$(157\text{ ft})\times\dfrac{1\text{ mi}}{5280\text{ ft}}\times\dfrac{1\text{ km}}{0.62\text{ mi}}=\dfrac{157}{5280\times0.62}\text{ km}\\\\\approx\boxed{0.0479594\text{ km}}$

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<em>Additional comment</em>

You can use Google to do units conversion by typing "3 feet to mm" or "157 feet to km". However, the conversion factor for the latter will be more exact than the one given in this problem. Hence the Google answer to the second question will not be applicable here.

7 0

2 years ago