Ask question

Ask question

All categories

3 years ago

7

Which expressions are equivalent to the one below? Check all that apply.

1 answer:

Paha777 [63]3 years ago

8 0

Answer:

The answer is C.

Step-by-step explanation:

You have to apply Logarithm Law,

$log( {a}^{n} ) \: ⇒ \: n log(a)$

So for this question :

$log( {10}^{3} ) = 3 log(10)$

You might be interested in

Where is 11/3 on a number line between 2 and 4

lions [1.4K]

Hiii! To work it out, you would have to do 11 ÷ 3. Which would equal 3.6 (Recurring). If it's to one decimal place, it would be 4 if not, it would just be 3.6 ^^

Hope it helps!! Good luck~

7 0

3 years ago

Help needed on this composition math problem

Marrrta [24]

Given that f(x) = x/(x - 3) and g(x) = 1/x and the application of <em>function</em> operators, f ° g (x) = 1/(1 - 3 · x) and the domain of the <em>resulting</em> function is any <em>real</em> number except x = 1/3.

<h3>How to analyze a composed function</h3>

Let be f and g functions. Composition is a <em>binary function</em> operation where the <em>variable</em> of the <em>former</em> function (f) is substituted by the <em>latter</em> function (g). If we know that f(x) = x/(x - 3) and g(x) = 1/x, then the <em>composed</em> function is:

$f\,\circ\,g \,(x) = \frac{\frac{1}{x} }{\frac{1}{x}-3}$

$f\,\circ\,g\,(x) = \frac{\frac{1}{x} }{\frac{1-3\cdot x}{x} }$

$f\,\circ\,g\,(x) = \frac{1}{1-3\cdot x}$

The domain of the function is the set of x-values such that f ° g (x) exists. In the case of <em>rational</em> functions of the form p(x)/q(x), the domain is the set of x-values such that q(x) ≠ 0. Thus, the domain of f ° g (x) is $\mathbb{R} - \{\frac{1}{3} \}$ .

To learn more on composed functions: brainly.com/question/12158468

#SPJ1

3 0

2 years ago

Please help me asap!!!

Deffense [45]

Answer:

8

Step-by-step explanation:

8 0

2 years ago

Given the functions: f(x) = x^2-9 and g(x) = 3 what is (f/g)(x)? A. x-3 B. x-9 C. x+3 D. x+9

uysha [10]

Answer:

f + g)(x) = f (x) + g(x)

= [3x + 2] + [4 – 5x]

= 3x + 2 + 4 – 5x

= 3x – 5x + 2 + 4

= –2x + 6

(f – g)(x) = f (x) – g(x)

= [3x + 2] – [4 – 5x]

= 3x + 2 – 4 + 5x

= 3x + 5x + 2 – 4

= 8x – 2

(f × g)(x) = [f (x)][g(x)]

= (3x + 2)(4 – 5x)

= 12x + 8 – 15x2 – 10x

= –15x2 + 2x + 8

\left(\small{\dfrac{f}{g}}\right)(x) = \small{\dfrac{f(x)}{g(x)}}(

g

f

)(x)=

g(x)

f(x)

= \small{\dfrac{3x+2}{4-5x}}=

4−5x

3x+2

My answer is the neat listing of each of my results, clearly labelled as to which is which.

( f + g ) (x) = –2x + 6

( f – g ) (x) = 8x – 2

( f × g ) (x) = –15x2 + 2x + 8

\mathbf{\color{purple}{ \left(\small{\dfrac{\mathit{f}}{\mathit{g}}}\right)(\mathit{x}) = \small{\dfrac{3\mathit{x} + 2}{4 - 5\mathit{x}}} }}

3 0

3 years ago

True or false. Tan^2 x = 1 - cos2x/ 1 + cos 2x

koban [17]

<u>ANSWER</u>

True

<u>EXPLANATION</u>

The given trigonometric equation is

$\tan^{2} (x) = \frac{1 - \cos(2x) }{1 + \cos(2x) }$

Recall the double angle identity:

$\cos(2x) = \cos^{2} x - \sin^{2}x$

We apply this identity to obtain:

$\tan^{2} (x) = \frac{1 - (\cos^{2} x - \sin^{2}x) }{1 + (\cos^{2} x - \sin^{2}x) }$

We maintain the LHS and simplify the RHS to see whether they are equal.

Expand the parenthesis

$\tan^{2} (x) = \frac{1 - \cos^{2} x + \sin^{2}x }{1 + \cos^{2} x - \sin^{2}x}$

$\implies\tan^{2} (x) = \frac{1 - \cos^{2} x + \sin^{2}x }{1 - \sin^{2}x + \cos^{2} x }$

Recall that:

$1 - \sin^{2}x = \cos^{2}x$

$1 - \cos^{2}x = \sin^{2}x$

We apply these identities to get:

$\implies\tan^{2} (x) = \frac{\sin^{2}x + \sin^{2}x }{\cos^{2} x + \cos^{2} x }$

$\implies\tan^{2} (x) = \frac{2\sin^{2}x }{ 2\cos^{2} x }$

$\implies\tan^{2} (x) = \frac{\sin^{2}x }{ \cos^{2} x }$

$\implies \tan^{2} (x) =( \frac{\sin x }{ \cos x })^{2}$

Also

$\frac{\sin x }{ \cos x } = \tan(x)$

$\implies \tan^{2} (x) =( \tan x )^{2}$

$\implies \tan^{2} (x) =\tan^{2} (x)$

Therefore the correct answer is True

5 0

3 years ago