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3 years ago

11

The functions f(x) and g(x) are defined below.

1 answer:

Vilka [71]3 years ago

8 0

So you do g of x first which is root of -4 and then sub that into f of x to get root minus 4 to the power of 4 - 2 root -4 squared + 2 which is the same as (((-4)^1/2)^2) - 4 times -4 + 2 which is the same as -4^2+16 +2 which is the same as 16+16+2 which is 34. Hope this helps :)

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A 20-gallon salt-water solution contains 15% pure salt. How much pure water should be added to it to produce a 10% solution?

Elanso [62]

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4(3x^2y^4)^3 / (2x^3y^5)^4

tatyana61 [14]

Solving the expressions $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$ we get $\frac{27}{4x^{6}y^{8}}$

Step-by-step explanation:

We need to Solve the expression: $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$

Solving:

$\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$

$=\frac{4(3^3x^6y^{12})}{(2^4x^{12}y^{20})}\\=\frac{4(27x^6y^{12})}{(16x^{12}y^{20})}\\=\frac{108x^6y^{12}}{16x^{12}y^{20}}$

Applying exponent rule: $\frac{x^a}{x^b}=x^{a-b}$

$=\frac{108x^{6-12}y^{12-20}}{16}\\=\frac{27x^{-6}y^{-8}}{4}$

Another exponent rule says: $x^{-a}=\frac{1}{x^a}=$

$=\frac{27}{4x^{6}y^{8}}$

So, solving the expressions $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$ we get $\frac{27}{4x^{6}y^{8}}$

Keywords: Solving Exponents

Learn more about Solving Exponents at:

#learnwithBrainly

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