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gladu [14]
3 years ago
12

1 1/45 into a improper fraction

Mathematics
1 answer:
balu736 [363]3 years ago
4 0
46/45. do u know how 2 do it? hope u do
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What is the value of the function y=1.3x+22 when x=-7?<br> y= -----------
rodikova [14]
y=1.3x+22

y=1.3(-7)+22

y = 12.9
5 0
3 years ago
Read 2 more answers
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
Help again please i need to pass this
Lena [83]

√50x³y

= √5²×2×x²×x×y

= 5x√2xy

d is the correct answer

8 0
3 years ago
Read 2 more answers
Solve the proportional equation below: 5/15 = 3/v
slavikrds [6]

Answer:

15

Step-by-step explanation:

5/15=3/v

5v=15×3

5v=75

5v/5=75/5

v=15

3 0
3 years ago
Read 2 more answers
A student created this table to represent a linear relationship between x and y.X Y-2. 10.0-1. 7.50. 5.01. 2.52. 0The student sa
alex41 [277]

The relationship between x and y is represent as:

Since, the relationship is linear.

The standard form of equation of line is:

y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)

Consider any two set x and y values from the given relationship.

Let (-2, 10) and (-1,7.5)

\text{Substitute x}_1=-2,y_1=10,x_2=-1,y_2=7.5\text{ in the standard equation of line.}

\begin{gathered} y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1) \\ y-10=\frac{7.5-10}{-1-(-2)}(x-(-2)) \\ y-10=\frac{-2.5}{1}(x+2) \\ y-10=-2.5(x+2) \\ y=-2.5(x+2)+10 \end{gathered}

The equation of the linear relationship between x and y is:

y = -2.5(x + 2) + 10

Now, to check that the point (9, -17.5) lies on the represented relationship between x and y

Substitute x = 9 and y = -17.5 in the equation y = -2.5(x + 2) + 10

y = -2.5(x + 2) + 10

-17.5 = -2.5(9 + 2) + 10

-17.5 = -2.5(11) + 10

-17.5 = -27.5 + 10

-17.5 = -17.5

Thus, LHS = RHS

Hence the point (9, -17.5) lie on the given linear relationship between x and y.

Answer: The point (9, -17.5) lie on the given linear relationship between x and y.

8 0
1 year ago
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