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Rashid [163]
2 years ago
11

6E and 7 with full explanation

Mathematics
2 answers:
ELEN [110]2 years ago
7 0

Answer:

See below ~

Step-by-step explanation:

Question 6(e) :

⇒ x² ≤ 16

<em>Take square root on each side :</em>

⇒ √x² ≤ √16

⇒ x ≤ 4 and x ≥ -4

⇒ -4 ≤ x ≤ 4

Setler [38]2 years ago
5 0
<h3><u>Part 6e</u></h3>

⇒ x² ≤ 16

Apply the <u>absolute rule</u> if x² < a then -√a < x < √a

⇒ -√16 ≤ x ≤ +√16

⇒ -4 ≤ x ≤ 4

<h3><u>Part 7</u></h3>

First of all, The student should have <u>subtracted both sides by 4</u>

What he should have done:

\rightarrow \sf \dfrac{3}{x-2} > 4

\rightarrow \sf \dfrac{3}{x-2} -4 > 4 -4

\rightarrow \sf \dfrac{3}{x-2} -\dfrac{4(x-2)}{x-2} > 0

\rightarrow \sf \dfrac{3-4(x-2)}{x-2}} > 0

\rightarrow \sf \dfrac{3-4x+8}{x-2}} > 0

\rightarrow \sf \dfrac{-4x+11}{x-2}} > 0

\sf \large \boxed{\sf {\left \{ {{-4x+11 = 0} \atop {x-2 = 0}} \right. }}

⇒ -4x + 11 = 0, x - 2 = 0

⇒ -4x = -11, x = 2

⇒ x = 11/4 , x = 2

Solution <u>satisfying the inequality</u>:

⇒ 2 < x < 11/4

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