We are asked to determine the value of a such that the function f(x) = ax^2 + 5 is fit for the point given (-1,2). In this case, we substitute 2 to y and -1 to x. The result is then 2 = a*(-1)^2 + 5 ; 2 = a + 5; a is then equal to -3
This equation has infinite solutions, any real number makes it true.
AEB = CED = 180 - 45 - 14 = 121 deg
EDC = 180 - 121 - 27 = 32 deg
So angle D is 32 degrees
Answer:
A quadratic equation can be written as:
a*x^2 + b*x + c = 0.
where a, b and c are real numbers.
The solutions of this equation can be found by the equation:

Where the determinant is D = b^2 - 4*a*c.
Now, if D>0
we have the square root of a positive number, which will be equal to a real number.
√D = R
then the solutions are:

Where each sign of R is a different solution for the equation.
If D< 0, we have the square root of a negative number, then we have a complex component:
√D = i*R

We have two complex solutions.
If D = 0
√0 = 0
then:

We have only one real solution (or two equal solutions, depending on how you see it)
Answer:
2c - 18x + 74/5
Step-by-step explanation:
<u>Step 1: Distribute
</u>
2(c + 7) - 18x + 4/5
(2 * c) + (2 * 7) - 18x + 4/5
<em>2c + 14 - 18x + 4/5
</em>
<u>Step 2: Combine like terms
</u>
2c + 14 - 18x + 4/5
<em>2c - 18x + 74/5
</em>
Answer: 2c - 18x + 74/5