Answer:
the 2nd at the bottom left
Here , we are provided with a table which shows 5 consecutive terms of an arithmetic sequence . But before solving further , let's recall that ;
The n'th term of a Arithmetic Sequence let's say it be
is given by ;
Where , <u>d</u> is the common difference
Now , here we are given with ;
We have to find the 2nd , 3rd and 4th term respectively ,
Now , by using the above formula , 5th term can be written as ;
Putting the values and transposing 1st term to RHS , we have ;
Now , as we got the common difference , so we can find out the missing terms now ;



Now



Also ,



Now , The given table can be written as ;

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Answer:
(3 ± √23 * i) /4
Step-by-step explanation:
To solve this, we can apply the Quadratic Equation.
In an equation of form ax²+bx+c = 0, we can solve for x by applying the Quadratic Equation, or x = (-b ± √(b²-4ac))/(2a)
Matching up values, a is what's multiplied by x², b is what's multiplied by x, and c is the constant, so a = 2, b = -3, and c = 4
Plugging these values into our equation, we get
x = (-b ± √(b²-4ac))/(2a)
x = (-(-3) ± √(3²-4(2)(4)))/(2(2))
= (3 ± √(9-32))/4
= (3 ± √(-23))/4
= (3 ± √23 * i) /4
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).
2x+6=x+3 this is the answer