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Natasha2012 [34]
3 years ago
8

QUICK EASY 25 POINTS!!

Mathematics
1 answer:
MAVERICK [17]3 years ago
7 0

Answer:

1. leave the radical symbol and don't try to convert it to a decimal form

2. The approximated value differs from the exact solution and doesn't give a true equation.

3. Depending on the case, if she needs a mathematical answer, the exact value should be used, but for more practical applications, the rounded decimal form would be more usable.

Step-by-step explanation:

The step by step solution to the equation:

4x^3=756\\x^3=\frac{756}{4} =189\\x=\sqrt[3]{189} \\x=\sqrt[3]{3^3*7} \\x=3*\sqrt[3]{7}

1.- Exact solution means that if in the final step when solving for x the value of  x^3 is not a perfect cube, one needs to leave it indicated as a radical expression (with the radical symbol).

in our case, the cubic root of 189 is not a perfect cube. The factor form of 189 is: 3^3*7, so there is a perfect cube factor (3^3), but the other factor (7) is a prime number. Therefore 3 can get out of the root, while 7 stays inside.

2.- The equation was solved above, in exact form. Now to solve it giving a decimal approximation, we use a calculator to find the cubic root of 7, which is an irrational number with infinite number of decimals, the first of which we type here: \sqrt[3]{7} = 1.91293118...

Therefore, the decimal approximation to the solving for x would be:

x=3*\sqrt[3]{7}=3*1,91293118...=5.73879...=5.74

Where we rounded to two decimals as requested.

When we replace the exact answer in the original expression, we get a perfect equality:

4*x^3=756\\4* (3\sqrt[3]{7} )^3=756\\4*3^3*(\sqrt[3]{7} )^3=756\\4*27*7=756\\\\756=756

While if we use the approximate answer, we get:

4*x^3=756\\\\4*(5.74)^3=756\\\\4*189.119224=756\\756.476896=756

which is NOT a true equality.

3.- I would stick with the idea of showing the exact answer as answer to the mathematical equation. but for a practical case (for example she needs to by some material as a result of her equation solving, it would be more practical to take the numerical approximation to the store, instead of a cubic root of a number.

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