Solve 7x/3 < 2 please

Mathematics

2 answers:

Mekhanik [1.2K]2 years ago

8 0

Answer:

$\large\boxed{x$

Step-by-step explanation:

$\dfrac{7x}{3}$

VLD [36.1K]2 years ago

5 0

Answer:

x<6/7

Step-by-step explanation:

Multiply by 3 from both sides of equation.

3*7x/3<2*3

Simplify.

7x<6

Divide by 7 from both sides of equation.

7x/7<6/7

Simplify, to find the answer.

x<6/7 is the correct answer.

I hope this helps you, and have a wonderful day!

You might be interested in

Can someone find the surface area please

lozanna [386]

To find the surface area of any object, the first step that you must do is find all shapes that are present in the object.

Here we have 2 triangles and 3 rectangles.

Knowing this, simply find the area of each shape.

Since the triangles are the same, one is facing up and the other down, they both would be 15.6 cm^2.

The rectangles would be 54 cm^2. as there are 3 ( 6 x 3 cm) ones.

Add the values up to have the surface area.

It’s 85.2 cm^2.

3 0

3 years ago

Read 2 more answers

Select all of the potential solution(s) of the equation 2log5x=log54.

Ilya [14]

X= -2 and x= 2 are the answers

8 0

2 years ago

Read 2 more answers

P(x)= 3x^3-5x^2-14x-4

nexus9112 [7]

$\displaystyle\\
P(x)=3x^3-5x^2-14x-4\\\\
D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\
\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\
3x^3-5x^2-14x-4=0\\\\
$

$\displaystyle\\
\text{Verification}\\\\
3x^3-5x^2-14x-4=\\\\
=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\
=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\
 =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\
\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\
\Longrightarrow~~~P(x)~\vdots~(3x+1)$

$\displaystyle\\
3x^3-5x^2-14x-4=0\\
~~~~~-5x^2 = x^2 - 6x^2\\
~~~~~-14x =-2x-12x \\
3x^3+x^2 - 6x^2-2x-12x-4=0\\
x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\
(3x+1)(x^2-2x -4)=0\\\\
\text{Solve: } x^2-2x -4=0\\\\
x_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm \sqrt{4+16}}{2}=\frac{2\pm \sqrt{20}}{2}=\frac{2\pm 2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\
x_1 =1+\sqrt{5}\\
x_2 =1-\sqrt{5}\\
\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}$

7 0

2 years ago

Answerrrrrrrrrrr plssssssssss

posledela

Answer:

No Solutions

Step-by-step explanation: