MATHEMATICS HIGH SCHOOL

Find a quadratic model for the set of values: (-2, -20), (0, -4), (4,-20) Show your work

Answers

Answer 1

Answer:

For this case, the quadratic function in its generic form is given by:

We must find the values of the coefficients.

For this, we evaluate the given points.

For (0, -4):

For (-2, -20):

For (4, -20):

Therefore, for the values of a and b we have the following system of equations:

Resolving graphically (see attached image) we have:

Then, the quadratic model is:

Answer:

a quadratic model for the set of values is:

$y = -2x ^ 2 + 4x - 4$

Answer 2

Answer: A quadratic function:
$y=ax^2+bx+c$

First, take the point (0,-4) and plug the values (x,y) into the equation:
$-4=a * 0^2+b * 0 +c \n-4=c$

So the equation is $y=ax^2+bx-4$ .

Now plug the values of the other two points into the equation and set up a system of equation:
$-20=a * (-2)^2+b * (-2)-4 \n-20=a * 4^2+b * 4-4 \n \n-20+4=4a-2b \n-20+4=16a+4b \n \n-16=4a-2b \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ | / 2 \n-16=16a+4b \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |/ 4 \n \n-8=2a-b \n\underline{-4=4a+b} \n-12=6a \n(-12)/(6)=a \na=-2 \n \n-8=2a-b \n-8=2 * (-2)-b \n-8=-4-b \n-8+4=-b \n-4=-b \nb=4$

The function is:
$\boxed{y=-2x^2+4x-4}$

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Not all rectangles have a. diagonals that bisect each other.
b. diagonals that are congruent.
c. four congruent sides.
d. consecutive angles that are supplementary.

Answers

I think the answer is d.

Use the substitution method to solve 3x-5y=-8 and x+2y=1

Answers

$\boxed{ \left \{ {{3x-5y=-8} \atop {x+2y=1}} \right.} \n \n x+2y=1\Rightarrow \boxed{x=1-2y} \n------------- \n \boxed{\left \{ {{3x-5y=-8} \atop {x=1-2y}} \right. } \n\n3(1-2y)-5y=-8 \n 3-6y-5y=-8\n\n3-11y=-8\n\n-11y=-8-3 \n \n -11y=-11y \n \n \boxed{y=1} \Rightarrow \boxed{x=-1}$

How to solve this
The minimum value of f is f×3=-5 and f×1=2

Answers

F ========== 2....................

Final answer:

The question provides two equations involving the variable f. By isolating f in these equations, we derive two possible values for f. The minimum value for f is the smaller of these two derived values, which is -5/3.

Explanation:

To solve this we can use the principle that if two different values of f multiply with different numbers to equal different constants, we can set up a system of equations to find those values of f.

The given equations are f×3=-5 and f×1=2. Let's denote f×1 as f₁ and f×3 as f₂.