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Elodia [21]
2 years ago
8

Find the value of 216m³ - 8n³ if 6m - 2n=0 and mn=12​

Mathematics
1 answer:
mestny [16]2 years ago
6 0

Answer:

0

Step-by-step explanation:

Given that,

6m - 2n=0

=>6m=2n

=>n=3m   [divide both side by 2]

& mn=12​

Now,

216c - 8n³

= 216m³- 8(3m)³

=216m³-  8 .27m³

=216m³- 216m³

=0

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Figure I and Figure II are similar figures,
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\frac{ST}{EF}=\frac{WR}{CD}

Thus the option H is the correct option.

To find the correct proportion we need to know about the properties of slimier polygon.

<h3>What are the properties of similar polygon?</h3>

Two polygons are said to be similar or congruent pentagon-

The corresponding angle of two pentagons are congruent.

The ratio of corresponding sides is same for all the sides.

The figure given in the problem of similar two polygons with six sides.

Rearrange the second figure as shown in figure.

As for the similar polygons, the ratio of corresponding sides is same for all the sides.

Thus the sides shown in the two rearranged figure must be same. Therefore,

\frac{ST}{EF}=\frac{WR}{CD}

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\frac{ST}{EF}=\frac{WR}{CD}

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Learn more about the similar polygon here;

brainly.com/question/771807

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2 years ago
The equation C = 5 9 F − 160 9 gives the relation between temperature readings in Celsius and Fahrenheit.
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Step-by-step explanation:

The question is wrong. The correct equation is :

C=\frac{5}{9}F-\frac{160}{9}

We know that the equation gives the relation  between temperature readings in Celsius and Fahrenheit.

Therefore, giving that we know the value in Fahrenheit ''F'' we can find the reading in Celsius ''C''. This define a function C(F) that depends of the variable ''F''.

So for the incise (a) we answer Yes, C is a function of F.

For (b) we need to find the mathematical domain of this function. Giving that we haven't got any mathematical restriction, the mathematical domain of the function are all real numbers.

Dom (C) = ( - ∞ , + ∞)

For (c) we know that the water in liquid state and at normal atmospheric pressure exists between 0 and 100 Celsius.

Therefore the range will be

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Now, we need to find the domain for this range. We do this by equaliting and finding the value for the variable ''F'' :

For C = 0 :

0=\frac{5}{9}F-\frac{160}{9} ⇒ F=32

And for C = 100 :

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Therefore, the domain as relating temperatures of water in its liquid state is

Dom (C) = (32,212)

For (d) we only need to replace in the equation by F=71 and find the value of C ⇒

C=\frac{5}{9}F-\frac{160}{9} ⇒

C=(\frac{5}{9})(71)-\frac{160}{9}

C=\frac{65}{3} ≅ 21.67

C(71) = 21.67 °C

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