Evaluate each function. 14) g(x)= 3x² - 1; Find g(-3)

Mathematics

2 answers:

adelina 88 [10]2 years ago

5 0

$g(x) =3x^2 -1\\\\g(-3)=3(-3)^2 -1 \\ \\~~~~~~~~~=3(9) -1 \\\\~~~~~~~~~=27-1\\\\~~~~~~~~~=26$

ololo11 [35]2 years ago

4 0

Answer:

g(26)

Step-by-step explanation:

substitute -3 for x.

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

$g(-3)=3(-3)^{2} -1\\g(-3)= 3(9)-1\\g(-3)= 27-1\\g(-3)=26$

You might be interested in

Bryce tried to solve an equation step by step. \qquad\begin{aligned} \dfrac83&=3\left(c+\dfrac53\right)\\\\ \\ \dfrac83&

sesenic [268]

Answer:

Bryce is wrong in step 1 because he did not distribute 3 over 5/3

Explanation

Given the steps taken by bryce as shown, we are to find where he made an error

$\qquad\begin{aligned} \dfrac83&=3\left(c+\dfrac53\right)\\\\ \\ \dfrac83&=3c+\dfrac53&\green{\text{Step } 1}\\\\ \\ 1&=3c&\blue{\text{Step } 2}\\\\ \\ \dfrac13&=c&\purple{\text{Step } 3}\\\\ \end{aligned}$

Given the expression;

$\dfrac83&=3\left(c+\dfrac53\right)\\\\ \\$

Step 1:Expand the bracket using the distributive law;

8/3 = 3c + 3(5/3)

<em>Simplify</em>

8/3 = 3c + 15/3

Step 2: Subtract 15/3 from both sides

8/3 - 15/3 = 3c+15/3-15/3

(8-15)/3 = 3c

-7/3 = 3c

Step 3: Multiply both sides by 1/3

-7/3 * 1/3 = 3c * 1/3

-7/9 = c

Swap

c = -7/9

From the calculation, we can see that Bryce is wrong in step 1 because he did not distribute 3 over 5/3 thereby making his solution incorrect

3 0

3 years ago

Express p in terms of q if

Irina-Kira [14]

$p=x^2+\dfrac{1}{x^2}\\\\p=\dfrac{x^4}{x^2}+\dfrac{1}{x^2}\\\\p=\dfrac{x^4+1}{x^2}\qquad(*)$

$q=x+\dfrac{1}{x}\\\\q=\dfrac{x^2}{x}+\dfrac{1}{x}\\\\q=\dfrac{x^2+1}{x}\qquad\text{square both sides}\\\\q^2=\left(\dfrac{x^2+1}{x}\right)^2\\\\q^2=\dfrac{(x^2+1)^2}{x^2}\qquad\text{use}\ \ (a+b)^2=a^2+2ab+b^2\\\\q^2=\dfrac{(x^2)^2+2(x^2)(1)+1^2}{x^2}\\\\q^2=\dfrac{x^4+2x^2+1}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+\dfrac{2x^2}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+2\qquad\text{subtract 2 from both sides}\\\\q^2-2=\dfrac{x^4+1}{x^2}\\\\\text{From (*) we have}\\\\\boxed{p=q^2-2}$