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cluponka [151]
3 years ago
14

There is anyone who r from India​

Mathematics
1 answer:
nataly862011 [7]3 years ago
7 0

Answer:i dont think so

Step-by-step explanation:

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Need Help! I don't get it
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Well hmmm let's say you take the car and go in the city for 60 miles with it, well, the car can do 60 miles per gallon, since you just drove it for 60 miles, you only spent 1 gallon of gasoline then.

that only happens if you drive it for 60 miles, what if you drive it for more, let's do a quick table on that,

\bf \begin{array}{ccll}&#10;miles&cost\\&#10;\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\&#10;60&3.6(1)\\\\&#10;&3.6\left( \frac{60}{60} \right)\\\\&#10;120&3.6(2)\\\\&#10;&3.6\left( \frac{120}{60} \right)\\\\&#10;180&3.6(3)\\\\&#10;&3.6\left( \frac{180}{60} \right)&#10;\end{array}

and so on, now let's check if you less than 60 miles,

\bf \begin{array}{ccll}&#10;miles&cost\\&#10;\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\&#10;40&3.6\left( \frac{40}{60} \right)\\\\&#10;20&3.6\left( \frac{20}{60} \right)\\\\&#10;10&3.6\left( \frac{10}{60} \right)&#10;\end{array}

so, if you divide the amount of miles driven, by 60, when you have driven it for 120 miles, 120/60 is just 2, and the cost is for 2 gallons, or 3.6 * 2, which is 7.2 bucks, for 180 miles is 180/60 or 3 gallons for 3.6 * 3 bucks, and so on.

now, what if you drive it instead for "m" miles?

\bf \begin{array}{ccll}&#10;miles&cost\\&#10;\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\&#10;m&3.6\left( \frac{m}{60} \right)\\\\&#10;\end{array}\implies c=3.6\left( \cfrac{m}{60} \right)\implies c=\cfrac{3.6m}{60}&#10;\\\\\\&#10;c=\cfrac{3.6}{60}m\implies c=0.06m
3 0
3 years ago
Factor using the identity: a3 + b3 = (a + b)(a2 – ab + b2) 8z3 + 27
vivado [14]

Answer:  this is our required factor i.e.

8z^3+27=(2z+3)(4z^2-6z+9)

Explanation:

Since we have given that

8z^3+27

As we know the identity , which says that

a^3+b^3=(a+b)(a^2-ab+b^2)

So, we can use this here ,

8z^3+27=(2z)^3+3^3\\\\=(2z+3)((2z)^2-2z\times 3+3^2)\\\\=(2z+3)(4z^2-6z+9)

Hence this is our required factor i.e.

8z^3+27=(2z+3)(4z^2-6z+9)


7 0
3 years ago
Read 2 more answers
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