A polynomial of four terms is sometimes called a quadrinomial, but there's really no need for such words.
Likewise, what is a 4th degree polynomial? Fourth degree polynomials are also known as quartic polynomials. Quartics have these characteristics: Zero to four roots. One, two or three extrema. Zero, one or two inflection points.
The degree of a polynomial is the highest of the degrees of the polynomial's monomials (individual terms) with non-zero coefficients. The degree of a term is the sum of the exponents of the variables that appear in it, and thus is a non-negative integer.
Zero Polynomial. The constant polynomial. whose coefficients are all equal to 0. The corresponding polynomial function is the constant function with value 0, also called the zero map. The zero polynomial is the additive identity of the additive group of polynomials.
Answer:
Step-by-step explanation:
Let's call hens h and ducks d. The first algebraic equation says that 6 hens (6h) plus (+) 1 duck (1d) cost (=) 40.
The second algebraic equations says that 4 hens (4h) plus (+) 3 ducks (3d) cost (=) 36.
The system is
6h + 1d = 40
4h + 3d = 36
The best way to go about this is to solve it by substitution since we have a 1d in the first equation. We will solve that equation for d since that makes the most sense algebraically. Doing that,
1d = 40 - 6h.
Now that we know what d equals, we can sub it into the second equation where we see a d. In order,
4h + 3d = 36 becomes
4h + 3(40 - 6h) = 36 and then simplify. By substituting into the second equation we eliminated one of the variables. You can only have 1 unknown in a single equation, and now we do!
4h + 120 - 18h = 36 and
-14h = -84 so
h = 6.
That means that each hen costs $6. Since the cost of a duck is found in the bold print equation above, we will sub in a 6 for h to solve for d:
1d = 40 - 6(6) and
d = 40 - 36 so
d = 4.
That means that each duck costs $4.
Answer:
2(-17+a^2+2g^2)
Step-by-step explanation:
step1
Answer:
Step-by-step explanation:
LOL The graph doesn’t match the y intercept :)
Anyway
If we have point (0,-3) we have a quadratic of
y=ax^2+bx-3 we are given points (-1,0) and (2,0) so
a-b-3=0 and 4a+2b-3=0
4a+2b-3+2(a-b-3)=0
4a+2b-3+2a-2b-6=0
6a-9=0
6a=9
a=1.5, since a-b=3
1.5-b=3
b=-1.5
y=1.5x^2-1.5x-3
You can determine midpoint by
x-midpoint = (x₁ + x₂)/2
y-midpoint = (y₁ + y₂)/2
Given from the question
x-midpoint = -6
x₁ = -7
y-midpoint = 5
y₁ = 8
Asked from the question
(x₂, y₂)
Solution
Find x₂, input the value of x₁ and x-midpoint to the formula
(x₁ + x₂)/2 = x-midpoint
(-7 + x₂)/2 = -6
-7 + x₂ = -6 × 2
-7 + x₂ = -12
x₂ = -12 + 7
x₂ = -5
Find y₂, input the value of y₁ and y-midpoint to the formula
(y₁ + y₂)/2 = 5
(8 + y₂)/2 = 5
8 + y₂ = 5 × 2
8 + y₂ = 10
y₂ = 10 - 8
y₂ = 2
Answer
P₂ = (-5,2)
