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3 years ago

Question 2:

Mathematics

2 answers:

arlik [135]3 years ago

5 0

Answer:

2: $g(f(x))=x^2 - 3$

3: x ≠ 3

Step-by-step explanation:

2 : Here the given functions,

$f(x) = x^2-----(1)$

$g(x) = x - 3----(2)$

$\because g(f(x)) = g(x^2)$ ( From equation (1) ),

$=x^2-3$ ( From equation (2) )

3 :

$h(x) = \frac{1}{x-3}$

Since, it is a rational function,

A rational function is defined for all real numbers except those for which,

Denominator = 0,

If x - 3 = 0

⇒ x = 3

So, Domain of h(x) = R - {3}

i.e., the domain restriction for h(x) is x ≠ 3

Anna [14]3 years ago

4 0

f(x) = x^2 and g(x) = x - 3.

To find f(g(x)) replace the x in f(x) by g(x).

f(g(x)) = (x - 3)^2

= x^2 - 6x + 9.

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the combined age of april and laura is 23 years. laura's age is two years more than half of april's age. what is laura's age

Kipish [7]

Laura's age is 9 years.

Solution:

Let x be the age of April.

Laura's age = 2 years more than half of April's age

<u>Convert statement into algebraic expression:</u>

Half of April's age = $\frac{x}{2}$

2 years more than half of April's age = $\frac{x}{2}+2$

Combined age of April and Laura = 23

⇒ April's age + Laura's age = 23

$$\Rightarrow \ \ x+(\frac{x}{2}+2) =23$

$$\Rightarrow \ \ x+\frac{x}{2}+2 =23$

$$\Rightarrow \ \ \frac{x}{1}+\frac{x}{2}+\frac{2}{1} =23$

To add the fractions make the denominators same.

Multiplying 2 on both numerator and denominator of unlike terms, we get

$$\Rightarrow \ \ \frac{x\times2}{1\times2}+\frac{x}{2}+\frac{2\times2}{1\times2} =23$

$$\Rightarrow \ \ \frac{2x}{2}+\frac{x}{2}+\frac{4}{2} =23$

Denominators are same, now add the fractions.

$$\Rightarrow \ \ \frac{2x+x+4}{2} =23$

Do cross multiplication.

$$\Rightarrow \ \ 2x+x+4=23\times2$

$$\Rightarrow \ \ 3x=46-4$

$$\Rightarrow \ \ 3x=42$

$$\Rightarrow \ \ x=14$

Aprils's age = 14 years

Laura's age = $\frac{14}{2}+2$

= 7 + 2

Laura's age = 9

Hence Laura's age is 9 years.

8 0

3 years ago

Are the functions f(x) = (x^2-1)/(x-1) and g(x)= x+1 equal for all x?

Vinvika [58]

$\bf \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-------------------------------\\\\$

$\bf f(x)=\cfrac{x^2-1}{x-1}\implies f(x)=\cfrac{x^2-1^2}{x-1}\implies f(x)=\cfrac{(\underline{x-1})(x+1)}{\underline{x-1}}
\\\\\\
f(x)=x+1\qquad \qquad \qquad \qquad \qquad g(x)=x+1\\\\
-------------------------------\\\\
\textit{they're, kinda, except that, when x = 1}
\\\\\\
g(x)=(1)+1\implies g(x)=2
\\\\\\
f(x)=\cfrac{(1)^2-1}{(1)-1}\implies f(x)=\cfrac{0}{0}\impliedby und efined$

7 0

3 years ago

I got stuck on this question.<br> Use long division to solve

Phantasy [73]

Answer:

5x² +19x +76 +310/(x-4)

Step-by-step explanation:

The process is straightforward. Find the quotient term, multiply it by the divisor and subtract from the dividend to get the new dividend. Repeat until the dividend is a constant (lower-degree than the divisor).

The tricky part with this one is realizing that there is no x-term in the original dividend, so that term needs to be added with a 0 coefficient. The rather large remainder is also unexpected, but that's the way this problem unfolds.

Unlike numerical long division, polynomial long division is simplified by the fact that the quotient term is the ratio of the highest-degree terms of the dividend and divisor. Here, the first quotient term is (5x^3)/(x) = 5x^2.