Answer:
12,8
Step-by-step explanation:
Just line them up and you will arrive at your answer:
 5,8
+ 7
_____
12,8
I am joyous to assist you anytime.
 
        
                    
             
        
        
        
Answer:
above is the solution to the question
 
        
             
        
        
        
I assume that the parabola in this particular problem is one whose axis of symmetry is parallel to the y axis. The formula we're going to use in this case is (x-h)2=4p(y-k). We know variables h and k from the vertex (1,20) but p is not given. However, we can solve for p by substituting values x and y in the formula with the y-intercept:
(0-1)^2=4p(16-20)
Solving for p, p=-1/16.
Going back to the formula, we can finally solve for the x-intercepts. Simply fill in variables p, h and k then set y to zero:
(x-1)^2=4(-1/16)(0-20)
(x-1)^2=5
x-1=(+-)sqrt(5)
x=(+-)sqrt(5)+1
Here, we have two values of x
x=sqrt(5)+1 and
x=-sqrt(5)+1
thus, the answers are: (sqrt(5)+1,0) and (-sqrt(5)+1,0).
        
             
        
        
        
Answer:
5x-5=3x +17 So were solving for x The answer would be =11
Step-by-step explanation:
 
        
                    
             
        
        
        
Answer:
(A)  with
 with  .
.
(B)  with
 with 
(C)  with
 with 
(D)  with
 with  ,
,
Step-by-step explanation
(A) We can see this as separation of variables or just a linear ODE of first grade, then  . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form
. With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form  with
 with  real.
 real. 
(B) Proceeding and the previous item, we obtain  . Which is not a vector space with the usual operations (this is because
. Which is not a vector space with the usual operations (this is because  ), in other words, if you sum two solutions you don't obtain a solution.
), in other words, if you sum two solutions you don't obtain a solution.
(C) This is a linear ODE of second grade, then if we set  and we obtain the characteristic equation
 and we obtain the characteristic equation  and then the general solution is
 and then the general solution is  with
 with  , and as in the first items the set of solutions form a vector space.
, and as in the first items the set of solutions form a vector space. 
(D) Using C, let be  we obtain that it must satisfies
 we obtain that it must satisfies  and then the general solution is
 and then the general solution is  with
 with  , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).
, and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).