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3 years ago

-6 + (-2) + 5= ? Help would really be appreciated!! :)

Mathematics

2 answers:

lozanna [386]3 years ago

6 0

Answer:

-6 + (-2) + 5 = -3

(Hope this helps! Btw, I am the first to answer. Brainliest pls! :D)

kvasek [131]3 years ago

4 0

Answer:

-3

Step-by-step explanation:

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Maya visits the movie rental store on one wall there are 6 DVDs on each of 5 shelves on another wall there are 4 DVDs on each of

guapka [62]

6 x 5 = 30
4 x 4 = 16
30 + 16 = 46
There are 46 DVDs in total.

4 0

3 years ago

Read 2 more answers

What does this mean -(-c) when c+2/3 on the numberline

ad-work [718]

Answer:

Step-by-step explanation:

a negative minus a negative is a positive so it mean c plus 2divided by 3 on the the number line

5 0

3 years ago

-3/v=-6 simply as much as possible

Aleksandr [31]

Hey!

In order to simplify this equation, we'll first have to multiply both sides of the equation by v. This will give us v on its own.

Original Equation :
$\frac{-3}{v} = -6$

New Equation {Added Multiply Both Sides by V} :
$\frac{-3}{v} v=-6v$

Solution {New Equation Solved} :
$-3 = -6v$

Now we'll switch sides to get v on the left side of the equation which is generally where we always want the variables to be located in these types of equations.

Old Equation :
$-3 = -6v$

New Equation {Switched} :
$-6v=-3$

Now we'll divide both sides by v to get v on its own.

Old Equation :
$-6v = -3$

New Equation {Added Divide Both Sides by V} :
$\frac{-6v}{v} = \frac{-3}{v}$

Solution {New Equation Solved} :
$v = \frac{1}{2}$

So, this means that in the equation $\frac{-3}{v} =-6$ , $v = \frac{1}{2}$ .

Hope this helps!

- Lindsey Frazier ♥

3 0

3 years ago

Write an equation of the line that passes through (4, −6) and is parallel to the line y=−3x−9.

Troyanec [42]

Answer:

Step-by-step explanation:

Strange question to me

6 0

3 years ago

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