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2 years ago

Simplify the expression

Mathematics

1 answer:

Jet001 [13]2 years ago

4 0

Answer: C

Step-by-step explanation:

48/64=3/4

a^(4-2)x b^(3-2)x c^(1-2)=a^2xb/c

You might be interested in

F(x)=3x 2 +9f, left parenthesis, x, right parenthesis, equals, 3, x, squared, plus, 9 and g(x)=\dfrac{1}{3}x^2-9g(x)= 3 1 x 2

34kurt

Answer:

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

$g(f(x)) = 3x^4 + 18x^2 + 18$

f(x) and g(x) and not inverse functions

Step-by-step explanation:

Given

$f(x) = 3x^2 + 9$

$g(x) = \dfrac{1}{3}x^2 - 9$

Required

Determine f(g(x))

Determine g(f(x))

Determine if both functions are inverse:

Calculating f(g(x))

$f(x) = 3x^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)(\frac{1}{3}x^2 - 9) + 9$

Expand Brackets

$f(g(x)) = (x^2 - 27)(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = x^2(\frac{1}{3}x^2 - 9) - 27(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = \frac{1}{3}x^4 - 9x^2 - 9x^2 + 243 + 9$

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

Calculating g(f(x))

$g(x) = \dfrac{1}{3}x^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)(3x^2 + 9) - 9$

$g(f(x)) = (x^2 + 3)(3x^2 + 9) - 9$

Expand Brackets

$g(f(x)) = x^2(3x^2 + 9) + 3(3x^2 + 9) - 9$

$g(f(x)) = 3x^4 + 9x^2 + 9x^2 + 27 - 9$

$g(f(x)) = 3x^4 + 18x^2 + 18$

Checking for inverse functions

$f(x) = 3x^2 + 9$

Represent f(x) with y

$y = 3x^2 + 9$

Swap positions of x and y

$x = 3y^2 + 9$

Subtract 9 from both sides

$x - 9 = 3y^2 + 9 - 9$

$x - 9 = 3y^2$

$3y^2 = x - 9$

Divide through by 3

$\frac{3y^2}{3} = \frac{x}{3} - \frac{9}{3}$

$y^2 = \frac{x}{3} - 3$

Take square root of both sides

$\sqrt{y^2} = \sqrt{\frac{x}{3} - 3}$

$y = \sqrt{\frac{x}{3} - 3}$

Represent y with g(x)

$g(x) = \sqrt{\frac{x}{3} - 3}$

Note that the resulting value of g(x) is not the same as $g(x) = \dfrac{1}{3}x^2 - 9$

Hence, f(x) and g(x) and not inverse functions

4 0

2 years ago

On the first january 2014 carol invested some money in a bank account the account payes 2.5% compound interest per year on 1st j

Sladkaya [172]

Answer:

$23,360

Step-by-step explanation:

Calculation to determine how much carol originally invested in the account

First step is to divide £23517.60 by 1.025

= (23,517.60)/(1+.025)

= (23,517.60)/1.025

=$22,944

Second step is to add back the $1,000 withdrew

=$22,944+$1,000

=$23,944

Now let calculate how much carol originally invested in the account

$23,944=1.025P

Divide both side by 1.025

P=$23,944/1.025

P=$23,360

Therefore the amount that carol originally invested in the account is $23,360

4 0

2 years ago

Which statements are true of the graph of h(x) = RootIndex 3 StartRoot x minus 4 EndRoot? Check all that apply. The domain of h(

Nastasia [14]

True statements:

The domain of h(x) is the set of all real numbers.

The range of h(x) is the set of all real numbers

Step-by-step explanation:

The function given in this problem is

$h(x)=\sqrt[3]{x-4}$

Now let's analyze the statements given:

- The domain of h(x) is the set of all real numbers. --> TRUE. The domain of a function is the set of values allowed for x. Here there is no restriction for the value of x, since the argument of an odd root can have any value (either positive or negative), so the domain of h(x) is the set of all real numbers.

- The range of h(x) is the set of all real numbers. --> TRUE. The range of a function is the set of values that the y can take. In this case, y can take any value: in fact, as $x \rightarrow +\infty$ , then $y \rightarrow +\infty$ , and as $x \rightarrow -\infty$ , then $y \rightarrow -\infty$ , and since there are no points of discontinuity in the function, this means that it can take any value of y.

For all points (x, h(x)), h(x) exists if and only if x – 4 > 0 --> FALSE. As we stated in part 1), the argument of a cubic root can be either positive or negative, so the function is defined also for x - 4 < 0.

The graph of h(x) is a translation of f(x) down 4 units. --> FALSE. The form of f(x) is not given so we can't evaluate this statement.

The graph of h(x) intercepts the x-axis at (4, 0). --> FALSE. In fact, when $x=0$ , the value of the function is