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3 years ago

Solve the inequality. 3x + 7 > 4x + 2

Mathematics

2 answers:

ankoles [38]3 years ago

7 0

5>x because you can subtract 5 from both sides and 3x from both sides

Nutka1998 [239]3 years ago

5 0

By solving this kind inequality, we would then have to find what does (x) equal, and then from here, we would determine which would be greater.

$rearrange: \\ 3*x+7-(4*x+2)\ \textgreater \ 0 \\ \\ Multiply \ both \ sides \ by \ (-1) \\ \\$

Then after, we add both sides by 5.

3x + 7= $3*5+7$ =22

4x + 2= $4*5+2$ =22

3x + 7 (x) 4x + 2

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How do i turn 7/3 in to a whole #

xenn [34]

Its 2 and 1/3
as a decimal its 2.3333
there is no possible way it can be written as a whole number

7 0

3 years ago

Read 2 more answers

The volume of a cylinder is 225π cubic inches,

aleksklad [387]

The value of the height of the cylinder is 25cm.

According to the statement

we have to find that the height pf the cylinder with the given value of the volume.

So, For this purpose we know that the

The volume of a cylinder is the density of the cylinder which finds that the amount of material it can carry. Cylinder's volume is given by the formula, πr^2h.

From the given information:

The volume of a cylinder is 225π cubic inches, and the radius of the cylinder is 3 inches.

Then

volume = πr^2h

225π = π3^2h

Now, solve it then

225 = 9h

h = 25.

The value becomes 25.

So, The value of the height of the cylinder is 25cm.

Learn more about volume of a cylinder here

brainly.com/question/9554871

#SPJ9

5 0

1 year ago

4-(2x-3)=3 whats the answer

Oxana [17]

Hello there.

Explanation:

↓↓↓↓↓↓↓↓↓↓

$4-(2x-3)=3$

First, subtract by 4 from both sides.

$4-(2x-3)-4=3-4$

Second, simplify.

$-(2x-3)=-1$

Third, divide by -1 from both sides.

$\frac{-(2x-3)}{-1}=\frac{-1}{-1}$

Fourth, simplify.

$2x-3=1$

Next, add by 3 from both sides.

$2x-3+3=1+3$

Then simplify.

$2x=4$

Therefore, divide by 2 from both sides.

$\frac{2x}{2}=\frac{4}{2}$

Finally, simplify.

$x=2$

Answer⇒⇒⇒⇒x=2

Hope this helps!

Thank you for posting your question at here on Brainly.

-Charlie

8 0

3 years ago

A number is greater than 8. The same number is less than 10. The inequalities x greater-than 8 and x less-than 10 represent the

jok3333 [9.3K]

Answer:

There are a few solutions because there are some fractions and decimals between 8 and 10

Step-by-step explanation:

Let the unknown number be 'x'

If the number is greater than 8 and the same number is less than 10, this can be expressed as;

x>8 and x < 10

Note that if x>8, then 8<x

The resulting inequalities are now;

8<x and x<10

Combining both inequalities we have: 8<x<10

Since the inequality didn't tell us that the variable 'x' is equal to 8 and 10, this means that our solution falls between 8 and 10 and the value of integer that falls within this range is 9. Other values that falls within this range are decimals and fractions.

Therefore it can be concluded that there are a few solutions because there are some fractions and decimals between 8 and 10

4 0

3 years ago

Read 2 more answers

Simplify u^2+3u/u^2-9 A.u/u-3, =/ -3, and u=/3 B. u/u-3, u=/-3

VashaNatasha [74]

  The correct answer is: Answer choice: [A]:
__________________________________________________________
→  " $\frac{u}{u-3}$ " ; " { u $\neq$ ± 3 } " ;

  →  or, write as: " u / (u − 3) " ;  {" u ≠ 3 "}  AND: {" u ≠ -3 "} ;
__________________________________________________________
Explanation:
__________________________________________________________
We are asked to simplify:

   $\frac{(u^2+3u)}{(u^2-9)}$ ;

Note that the "numerator" —which is: "(u² + 3u)" — can be factored into:
  → " u(u + 3) " ;

And that the "denominator" —which is: "(u² − 9)" — can be factored into:
  → "(u − 3) (u + 3)" ;
___________________________________________________________
Let us rewrite as:
___________________________________________________________

→ $\frac{u(u+3)}{(u-3)(u+3)}$ ;

___________________________________________________________

→ We can simplify by "canceling out" BOTH the "(u + 3)" values; in BOTH the "numerator" AND the "denominator" ;  since:

" $\frac{(u+3)}{(u+3)} = 1$ " ;

→ And we have:
_________________________________________________________

→  " $\frac{u}{u-3}$ " ; that is: " u / (u − 3) " ; { u $\neq 3$ } .
and: { u $\neq-3$ } .

→ which is:  "Answer choice: [A] " .
_________________________________________________________

NOTE:  The "denominator" cannot equal "0" ; since one cannot "divide by "0" ;

and if the denominator is "(u − 3)" ; the denominator equals "0" when "u = -3" ; as such:

"u $\neq$ 3" ;

→ Note: To solve: "u + 3 = 0" ;

Subtract "3" from each side of the equation;

→ " u + 3 − 3 = 0 − 3 " ;

→ u = -3 (when the "denominator" equals "0") ;

→ As such: " u $\neq$ -3 " ;

Furthermore, consider the initial (unsimplified) given expression:

→   $\frac{(u^2+3u)}{(u^2-9)}$ ;

Note:  The denominator is: "(u²  − 9)" .

The "denominator" cannot be "0" ; because one cannot "divide" by "0" ;

As such, solve for values of "u" when the "denominator" equals "0" ; that is:
_______________________________________________________

→ " u² − 9 = 0 " ;

→ Add "9" to each side of the equation ;

→ u² − 9 + 9 = 0 + 9 ;

→ u² = 9 ;

Take the square root of each side of the equation;
to isolate "u" on one side of the equation; & to solve for ALL VALUES of "u" ;

→ √(u²) = √9 ;

→ | u | = 3 ;

→ " u = 3" ; AND; "u = -3 " ;

We already have: "u = -3" (a value at which the "denominator equals "0") ;

We now have "u = 3" ; as a value at which the "denominator equals "0");

→ As such: " u $\neq 3$ " ; "u $\neq$ -3 " ;

or, write as: " { u $\neq$ ± 3 } " .

_________________________________________________________

6 0

3 years ago