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zysi [14]
2 years ago
8

Can anyone pls help me with this I'll mark as brainlist for correct answer please do explain in detail each answers Ty! (I'm pre

paring for my maths exam tmr)​

Mathematics
1 answer:
densk [106]2 years ago
7 0

<u>Question 8</u>

a^2 + 7a + 12

= (a+3)(a+4)

When factorising a quadratic, the product of the two factors should equal the constant term (12), and the sum of the two factors should equal the linear term (7). To find the two factors, list out the factors of 12 (1x12, 2x6, 3x4) and identify the pair that adds up to 7 (3+4).

An alternative method if you get stuck during your exam would be to solve it algebraically using the quadratic formula and then write it in the factorised form.

a = (-7 +or- sqrt(7^2 - 4(1)(12)) / 2(1)

= (-7 +or- sqrt(1))/2

= -3 or -4

These factors are the negative of the values that would go in the brackets when written in factorised form, as when a = -3 the factor (a+3) would equal 0. (If it were positive 3 instead, then in the factorised form it would be a-3).

<u>Question 10</u>

-3(x - y)/9 + (4x - 7y)/2 - (x + y)/18

Rewrite each fraction with a common denominator so you can combine the fractions into one.

= -6(x - y)/18 + 9(4x - 7y)/18 - (x + y)/18

= (-6(x - y) + 9(4x - 7y) - (x + y)) /18

Expand the brackets and collect like terms.

= (-6x + 6y + 36x - 63y - x - y)/18

= (29x - 58y)/18

= 29/18 x - 29/9 y

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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
What the error to this promblem
dimaraw [331]
The error to that problem is the person switched the x and y points, like in point A, it’s supposed to be (x,y) or (1,5). The person put the y in front of the x while writing the points.
3 0
3 years ago
a family's monthly utility costs are phone 200 electric 100 Cable TV 100 water and gas 50 internet 50 by eliminating cable by wh
AleksandrR [38]

Answer:

500

Step-by-step explanation:

I'm not sure

6 0
3 years ago
Y=mx+b (-6,-3) and (0,9)
Leno4ka [110]

Answer:

y = 2x + 9

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + b ( m is the slope and b the y- intercept )

Calculate m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (- 6, - 3) and (x₂, y₂ ) = (0, 9)

m = \frac{9+3}{0+6} = \frac{12}{6} = 2

The line crosses the y- axis at (0, 9) ⇒ b = 9

y = 2x + 9 ← equation of line

3 0
3 years ago
what is the answer to Lakshmi planted 20 begonias, but her neighbor’s dog ate 7 of them. What percent of the begonias did the do
IrinaK [193]

Answer: 35%

Step-by-step explanation:

7/20 * 100

= 0.35 * 100

= 35%

5 0
3 years ago
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