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aleksandrvk [35]
3 years ago
14

Which can be used to describe the expression? Check all that apply.

Mathematics
2 answers:
frozen [14]3 years ago
8 0
We can write the expression above as the following function:

f(r)=(r-4)^{3}

So let's examine the expressions that are true for this exercise.

1. <span>There are three factors of r-4:

This is true because we can write the function f(r) as follows:

</span>f(r)=(r-4)(r-4)(r-4)

So you can see that in fact there are three factors.

<span>2. The expression is equal to 1 over 12 factors of r.

This is false. It is obvious that this is impossible. There is no any way to get the same expression by applying this statement.
</span>
3. <span>Adding the exponents will create an equivalent expression.

This is true because we can write the function as follows:

</span>f(r)=(r-4)^{1+1+1}

So adding one three times we can get the same function, that is:

1+1+1=3

Therefore this is an equivalent expression because:

f(r)=(r-4)^{1+1+1}=(r-4)^{3}

4. Multiplying the exponents will create an equivalent expression.

This is true.You can get the following expression:

f(r)=(r-4)^{\frac{2}{3}\times \frac{9}{2}}

By multiplying the exponents we have:

\frac{2}{3}\times \frac{9}{2}=3

Therefore this is an equivalent expression because:

f(r)=(r-4)^{\frac{2}{3}\times \frac{9}{2}}=(r-4)^{3}

5. <span>The expression simplifies to.

The expression is simplified, that is, it has been factorized. Therefore there is no a way to simplify this function but:

</span>f(r)=(r-4)^{3}

Irina-Kira [14]3 years ago
8 0

A, B, D are Correct because i took the test

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find a curve that passes through the point (1,-2 ) and has an arc length on the interval 2 6 given by 1 144 x^-6
taurus [48]

Answer:

f(x) = \frac{6}{x^2} -8 or f(x) = -\frac{6}{x^2} + 4

Step-by-step explanation:

Given

(x,y) = (1,-2) --- Point

\int\limits^6_2 {(1 + 144x^{-6})} \, dx

The arc length of a function on interval [a,b]:  \int\limits^b_a {(1 + f'(x^2))} \, dx

By comparison:

f'(x)^2 = 144x^{-6}

f'(x)^2 = \frac{144}{x^6}

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f'(x) =\± \sqrt{\frac{144}{x^6}}

f'(x) = \±\frac{12}{x^3}

Split:

f'(x) = \frac{12}{x^3} or f'(x) = -\frac{12}{x^3}

To solve fo f(x), we make use of:

f(x) = \int {f'(x) } \, dx

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f(x) = \int {\frac{12}{x^3} } \, dx

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f(x) = \frac{12}{2x^2} + c

f(x) = \frac{6}{x^2} + c

We understand that it passes through (x,y) = (1,-2).

So, we have:

-2 = \frac{6}{1^2} + c

-2 = \frac{6}{1} + c

-2 = 6 + c

Make c the subject

c = -2-6

c = -8

f(x) = \frac{6}{x^2} + c becomes

f(x) = \frac{6}{x^2} -8

For: f'(x) = -\frac{12}{x^3}

f(x) = \int {-\frac{12}{x^3} } \, dx

Integrate:

f(x) = -\frac{12}{2x^2} + c

f(x) = -\frac{6}{x^2} + c

We understand that it passes through (x,y) = (1,-2).

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-2 = -\frac{6}{1^2} + c

-2 = -\frac{6}{1} + c

-2 = -6 + c

Make c the subject

c = -2+6

c = 4

f(x) = -\frac{6}{x^2} + c becomes

f(x) = -\frac{6}{x^2} + 4

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