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Over [174]
3 years ago
5

A box of oranges weighs 4kg. if one orange weighs 80 g how many oranges are in the box?

Mathematics
2 answers:
wlad13 [49]3 years ago
5 0

Answer:

50 Oranges is your answer.

Step-by-step explanation:

Lets see..

4 kg = 4000 grams mainly because 1kg = 1000 grams in total.

4000 grams divided by 80 grams will give you the number of oranges that the box has since each orange weighs 80 grams.

4000/80 = 50.

hope this helps!

7nadin3 [17]3 years ago
4 0

Answer:

50 Oranges

Step-by-step explanation:

1 kilogram = 1000 grams so if each orange weighs 80 grams 1,000 divided by 80 is 12.5 oranges. But there's 4 kilograms so if you do 12.5 times 4 it gives you 50 oranges which is your answer.

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Given that the initial dimension of the square garden is x, the initial area would become x². When any of the dimension is increased by 35 inches then, the area is increased by 35x. The statement that describes 35x is the additional area of the figure after it was extended for 35 inches. 
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3 years ago
Please someone help me...​
laiz [17]

Step-by-step explanation:

First factor out the negative sign from the expression and reorder the terms

That's

\frac{1}{ - (( \tan(2A) -  \tan(6A)  )}  -  \frac{1}{ \cot(6A)  -  \cot(2A) }

<u>Using trigonometric </u><u>identities</u>

That's

<h3>\cot(x)  =  \frac{1}{ \tan(x) }</h3>

<u>Rewrite the expression</u>

That's

\frac{1}{ - (( \tan(2A) -  \tan(6A)  )} -    \frac{1}{ \frac{1}{ \tan(6A) } }  -  \frac{1}{ \frac{1}{ \tan(2A) } }

We have

<h3>-  \frac{1}{  \tan(2A) -  \tan(6A)  } -   \frac{1}{ \frac{ \tan(2A) -  \tan(6A)  }{ \tan(6A) \tan(2A)  } }</h3>

<u>Rewrite the second fraction</u>

That's

<h3>-  \frac{1}{  \tan(2A) -  \tan(6A)  } -   \frac{ \tan(6A)  \tan(2A) }{ \tan(2A) -  \tan(6A)  }</h3>

Since they have the same denominator we can write the fraction as

-  \frac{1 +  \tan(6A) \tan(2A)  }{ \tan(2A) -  \tan(6A)  }

Using the identity

<h3>\frac{x}{y}  =  \frac{1}{ \frac{y}{x} }</h3>

<u>Rewrite the expression</u>

We have

<h3>-  \frac{1}{ \frac{ \tan(2A)  -  \tan(6A) }{1 +  \tan(6A) \tan(2A)  } }</h3>

<u>Using the trigonometric identity</u>

<h3>\frac{ \tan(x) -  \tan(y)  }{1 +  \tan(x)  \tan(y) }  =  \tan(x - y)</h3>

<u>Rewrite the expression</u>

That's

<h3>- \frac{1}{ \tan(2A -6A) }</h3>

Which is

<h3>-  \frac{1}{ \tan( - 4A) }</h3>

<u>Using the trigonometric identity</u>

<h3>\frac{1}{ \tan(x) }  =  \cot(x)</h3>

Rewrite the expression

That's

<h3>-  \cot( - 4A)</h3>

<u>Simplify the expression using symmetry of trigonometric functions</u>

That's

<h3>- ( -  \cot(4A) )</h3>

<u>Remove the parenthesis </u>

We have the final answer as

<h2>\cot(4A)</h2>

As proven

Hope this helps you

6 0
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Answer:

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Step-by-step explanation:

volume= \frac{2}{3}(l×w×h)

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4 years ago
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