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stich3 [128]
3 years ago
9

Find the solution set of the inequality \qquad4x -1 < 11.4x−1<11.4, x, minus, 1, is less than, 11, point \qquad xxx

Mathematics
2 answers:
Annette [7]3 years ago
7 0

Answer:

<em> (∞, 3]</em>

Step-by-step explanation:

Given the inequality 4x−1<11

We are to find the solution set;

4x−1<11

Add 1 to both sides

4x−1+1<11+1

4x < 12

Divide both sides by 4

4x/4 = 12/4

x < 3

<em>Hence the solution set of the inequality is (∞, 3]</em>

vesna_86 [32]3 years ago
5 0

Answer:

x > 5.

Step-by-step explanation:

1 / 3

Let’s start by subtracting \blue{14}14start color #6495ed, 14, end color #6495ed from both sides of the inequality:

\qquad\begin{aligned} 14\blue{-14} - 3x &< -1 \blue{-14}\\ - 3x &< -15 \\ \end{aligned}  

14−14−3x

−3x

​  

 

<−1−14

<−15

​  

 

Hint #22 / 3

Next, let's divide both sides by \green{-3}−3start color #28ae7b, minus, 3, end color #28ae7b. When you divide an inequality by a negative number, the inequality sign must be \text{\pink{reversed}}reversedstart text, start color #ff00af, r, e, v, e, r, s, e, d, end color #ff00af, end text:

\qquad\begin{aligned} - 3x& < -15\\ \\ \dfrac{-3x}{\green{-3}} &\pink{>} \dfrac{-15}{\green{-3}} \\ \\ x &\pink{>} 5\\ \end{aligned}  

−3x

−3

−3x

​  

 

x

​  

 

<−15

>  

−3

−15

​  

 

>5

​  

 

Hint #33 / 3

The solution set of the inequality is:

\qquad x > 5.

x>5.

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Step-by-step explanation:

1. y=(x+3)^3

y=0\\ (x+3)^3=0\\ \sqrt[3]{(x+3)^3}=\sqrt[3]{0}\\ x+3=0\\ x+3-3=0-3\\ x=-3

Zero: x=-3 multiplicity 3.


2. y=(x-2)^2 (x-1)

y=0\\ (x-2)^2(x-1)=0\\ \left \{ {{(x-2)^2=0} \atop {x-1=0}} \right\\ \left \{ {{\sqrt{(x-2)^2} =\sqrt{0} } \atop {x-1+1=0+1}} \right\\ \left \{ {{x-2=0} \atop {x=1}} \right\\ \left \{ {{x-2+2=0+2} \atop {x=1}} \right\\ \left \{ {{x=2} \atop {x=1}} \right.

Zeros: x=2 multiplicity 2; x=1 multiplicity 1


3. y=(2x+3)(x-1)^2

y=0\\ (2x+3)(x-1)^2=0\\ \left \{ {{2x+3=0} \atop {(x-1)^2=0}} \right\\ \left \{ {{2x+3-3=0-3} \atop {\sqrt{(x-1)^2} =\sqrt{0} }} \right\\ \left \{ {{2x=-3} \atop {x-1=0}} \right\\ \left \{ {{\frac{2x}{2} =\frac{-3}{2} } \atop {x-1+1=0+1}} \right\\ \left \{ {{x=-\frac{3}{2} } \atop {x=1}} \right.

Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.

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