There will be many polynomials of degree 2 that pass through the points (1, 5) and (3, 7). The situation can be described by a system of two linear equations in three variables that has many solutions. Find an equation (involving a parameter r) that represents this family of polynomials. (Let the coefficient of the x2 term in the equation be r.) y(x) = Determine the polynomials that open up and those that open down.

Step-by-step explanation:

Given that a quadratic function(polynomial of degree 2) passes through two points (1,5) and (3,7)

The function would be of the form

with if a>0 open up and if a<0 open down

Since passes through(1,5) and (3,7) these points satisfy the equation

Let us eliminate c easily

2 = 8a +2b or 4a+b =1

b =1-4a

Thus parametric equation we can write as

where c is arbitrary

If a>0 this will be open up otherwise open down.

no solution

step-by-step explanation:

there is not enough information in the problem to solve.

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.