Given that k(x) = (2x + 4)2

Mathematics

1 answer:

taurus [48]2 years ago

3 0

Answer:

$f(x)=4x\\g(x)=x+2$

Step-by-step explanation:

Because we have to rewrite this equation in the format $f(g(x))$ , we have to divide, or factor to find basic terms,

Expanding the value of k(x), we have $k(x)=4x+8$ . We see that each term can be divisible by 4, so we can factor out 4 to get

$k(x)=4(x+2)$

Now, we have two different terms getting multiplied. We can separate the two to get $4 \text{ and} (x+2)$

Because we are multiplying 4 by the other term, this is represented by $4x$

Now, we can just set f(x) and g(x) to these functions:

$f(x)=4x\\g(x)=x+2$

Now, just to make sure, we can plug a value into k(x) and the same value into f(g(x)). Plugging in 1, we have (2(1)+4)2 as 2(2+4), which is 2(6) = 12.

Plugging 1 into f(g(x)), we can evaluate g(1) first, to get 1 + 2 = 3. Now, f(3) = 4(3)= 12, which is the same for k(x).

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Katyanochek1 [597]

Answer:

b/a
16a²b²
n¹⁰/(16m⁶)
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m⁷n³n/m

Step-by-step explanation:

These problems make use of three rules of exponents:

$a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}$

In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)

1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.

$\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}$

2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.

$\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2$

3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.

$\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}$

4. This works the same way the previous problem does.

$\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}$

5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.

$\dfrac{m^7n^3}{mn^{-1}}=\dfrac{m^7n^3n}{m}$