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nika2105 [10]
3 years ago
6

- 100 points -

Mathematics
2 answers:
Molodets [167]3 years ago
8 0

Answer:

B

Step-by-step explanation:

mezya [45]3 years ago
4 0

D) - y = (x + 5)(x - 3)(x + 1).

<h3>EXPLANATION:</h3>

Table in this case would look like this:

Write coefficients of x³, x² ,x and the constant in a row and divisor would be the x value obtained by equation x + 5 = 0.

The sequence of multiplications would be as shown in picture.

x² - 2x - 3

x² - 3x + x - 3

x(x - 3) + 1(x - 3)

(x + 1)(x - 3)(x + 5).

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arlik [135]

Answer:

I think the answer is 60

Step-by-step explanation:

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7 0
3 years ago
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wlad13 [49]

it is equal to 2 1/2

Step-by-step explanation:

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What are the next 3 terms to this sequence?<br> 2, 6, − 4, 10, − 14, − 24...
Thepotemich [5.8K]

Answer:

-38, 62, -100

Step-by-step explanation:

I think you meant  a +24,  not the -24.

The recursive formula is   a_n = a_(n-2)  -  a_(n-1)

so. we try this 3 times after the -24 term

a_3 = -4 =  2 - 6 =   a_1 - a_2

...

a_6  = a_4 - a_5 = 10 - (-14) = 24

a_7 = a_5 - a_6 = -14 - (24) = -38

a_8 = a_6 - a_7 = 24 - (-38)  = 62

a_9 = a_7 - a_8 =  -38 - 62 = -100

a_10 = a_8 - a_9 = 62 - (-100) = 162

8 0
3 years ago
99 students choose to attend one of three after school activities: football, tennis or running.
nordsb [41]

Answer:   \dfrac{49}{99}

<u>Step-by-step explanation:</u>

There are 61 boys --> 38 girls

A) 19 chose football: 5 are girls --> 14 are boys

B) 38 girls: 28 chose running, 5 chose football --> 5 girls chose tennis

C) 31 students chose tennis: 5 are girls --> 26 are boys.

D) 61 boys: 14 chose football, 26 chose tennis --> 21 boys chose running.

\large\boxed{\begin{array}{l|cc||c}&\underline{Boys}&\underline{Girls}&\underline{Total}\\Football&14&5&19\\Tennis&26&5&31\\\underline{Running}&\underline{\quad 21\quad}&\underline{\quad 28\quad}&\underline{\quad 29\quad}\\Total&61&38&99\end{array}}

Total running = 21 boys + 28 girls = 49

Total students = 99

\dfrac{\text{Total running}}{\text{Total students}}=\large\boxed{\dfrac{49}{99}}

3 0
3 years ago
This 1 seems really complicated
Fofino [41]
The solution to this system set is:  "x = 4" , "y = 0" ;  or write as:  [4, 0] .
________________________________________________________
Given: 
________________________________________________________
 y = - 4x + 16 ; 

 4y − x + 4 = 0 ;
________________________________________________________
"Solve the system using substitution" .
________________________________________________________
First, let us simplify the second equation given, to get rid of the "0" ; 

→  4y − x + 4 = 0 ; 

Subtract "4" from each side of the equation ; 

→  4y − x + 4 − 4 = 0 − 4 ;

→  4y − x = -4 ;
________________________________________________________
So, we can now rewrite the two (2) equations in the given system:
________________________________________________________
   
y = - 4x + 16 ;   ===> Refer to this as "Equation 1" ; 

4y − x =  -4 ;     ===> Refer to this as "Equation 2" ; 
________________________________________________________
Solve for "x" and "y" ;  using "substitution" :
________________________________________________________
We are given, as "Equation 1" ;

→  " y = - 4x + 16 " ;
_______________________________________________________
→  Plug in this value for [all of] the value[s] for "y" into {"Equation 2"} ;

       to solve for "x" ;   as follows:
_______________________________________________________
Note:  "Equation 2" :

     →  " 4y − x =  - 4 " ; 
_________________________________________________
Substitute the value for "y" {i.e., the value provided for "y";  in "Equation 1}" ;
for into the this [rewritten version of] "Equation 2" ;
→ and "rewrite the equation" ;

→   as follows:  
_________________________________________________

→   " 4 (-4x + 16) − x = -4 " ;
_________________________________________________
Note the "distributive property" of multiplication :
_________________________________________________

   a(b + c)  = ab + ac ;   AND: 

   a(b − c) = ab <span>− ac .
_________________________________________________
As such:

We have:  
</span>
→   " 4 (-4x + 16) − x = - 4 " ;
_________________________________________________
AND:

→    "4 (-4x + 16) "  =  (4* -4x) + (4 *16)  =  " -16x + 64 " ;
_________________________________________________
Now, we can write the entire equation:

→  " -16x + 64 − x = - 4 " ; 

Note:  " - 16x − x =  -16x − 1x = -17x " ; 

→  " -17x + 64 = - 4 " ;   Solve for "x" ; 

Subtract "64" from EACH SIDE of the equation:

→  " -17x + 64 − 64 = - 4 − 64 " ;   

to get:  

→  " -17x = -68 " ;

Divide EACH side of the equation by "-17" ; 
   to isolate "x" on one side of the equation; & to solve for "x" ; 

→  -17x / -17 = -68/ -17 ; 

to get:  

→  x = 4  ;
______________________________________
Now, Plug this value for "x" ; into "{Equation 1"} ; 

which is:  " y = -4x + 16" ; to solve for "y".
______________________________________

→  y = -4(4) + 16 ; 

        = -16 + 16 ; 

→ y = 0 .
_________________________________________________________
The solution to this system set is:  "x = 4" , "y = 0" ;  or write as:  [4, 0] .
_________________________________________________________
Now, let us check our answers—as directed in this very question itself ; 
_________________________________________________________
→  Given the TWO (2) originally given equations in the system of equation; as they were originally rewitten; 

→  Let us check;  

→  For EACH of these 2 (TWO) equations;  do these two equations hold true {i.e. do EACH SIDE of these equations have equal values on each side} ; when we "plug in" our obtained values of "4" (for "x") ; and "0" for "y" ??? ; 

→ Consider the first equation given in our problem, as originally written in the system of equations:

→  " y = - 4x + 16 " ;    

→ Substitute:  "4" for "x" and "0" for "y" ;  When done, are both sides equal?

→  "0 = ?  -4(4) + 16 " ?? ;   →  "0 = ? -16 + 16 ?? " ;  →  Yes!  ;

 {Actually, that is how we obtained our value for "y" initially.}.

→ Now, let us check the other equation given—as originally written in this very question:

→  " 4y − x + 4 = ?? 0 ??? " ;

→ Let us "plug in" our obtained values into the equation;

 {that is:  "4" for the "x-value" ; & "0" for the "y-value" ;  

→  to see if the "other side of the equation" {i.e., the "right-hand side"} holds true {i.e., in the case of this very equation—is equal to "0".}.

→    " 4(0)  −  4 + 4 = ? 0 ?? " ;

      →  " 0  −  4  + 4 = ? 0 ?? " ;

      →  " - 4  + 4 = ? 0 ?? " ;  Yes!
_____________________________________________________
→  As such, from "checking [our] answer (obtained values)" , we can be reasonably certain that our answer [obtained values] :
_____________________________________________________
→   "x = 4" and "y = 0" ;  or; write as:  [0, 4]  ;  are correct.
_____________________________________________________
Hope this lenghty explanation is of help!  Best wishes!
_____________________________________________________
7 0
3 years ago
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