The degree of a polynomial is the highest power of its terms.
The power of a term is the sum of the powers of all the variables in a term.
A polynomial is written starting with the greatest power in standard form.
In the first case, the power of the first term is 3, the power of the second is 3 (2 from x + 1 from y) but the power of x has decreased so it is the second term, and then so on.
In the second case, the power is starting form 2 and then increasing to 3. This is incorrect.
Therefore, Marcus' suggestion is correct.
Answer:
Step-by-step explanation:
B(2,10); D(6,2)
Midpoint(x1+x2/2, y1+y2/2) = M ( 2+6/2, 10+2/2) = M(8/2, 12/2) = M(4,6)
Rhombus all sides are equal.
AB = BC = CD =AD
distance = √(x2-x1)² + (y2- y1)²
As A lies on x-axis, it y-co ordinate = 0; Let its x-co ordinate be x
A(X,0)
AB = AD
√(2-x)² + (10-0)² = √(6-x)² + (2-0)²
√(2-x)² + (10)² = √(6-x)² + (2)²
√x² -4x +4 + 100 = √x²-12x+36 + 4
√x² -4x + 104 = √x²-12x+40
square both sides,
x² -4x + 104 = x²-12x+40
x² -4x - x²+ 12x = 40 - 104
8x = -64
x = -64/8
x = -8
A(-8,0)
Let C(a,b)
M is AC midpoint
(-8+a/2, 0 + b/2) = M(4,6)
(-8+a/2, b/2) = M(4,6)
Comparing;
-8+a/2 = 4 ; b/2 = 6
-8+a = 4*2 ; b = 6*2
-8+a = 8 ; b = 12
a = 8 +8
a = 16
Hence, C(16,12)
Answer:
Step-by-step explanation:
<u><em>Given Equation is </em></u>
=>
Comparing it with , we get
=> a = 2, b = 7 and c = -9
So,
Sum of roots = α+β =
α+β = -7/2
Product of roots = αβ = c/a
αβ = -9/2
<em>Now, Finding the equation whose roots are:</em>
α/β ,β/α
Sum of Roots =
Sum of Roots =
Sum of Roots =
Sum of roots =
Sum of roots =
Sum of Roots =
Sum of roots =
Sum of roots = S =
Product of Roots =
Product of Roots = P = 1
<u><em>The Quadratic Equation is:</em></u>
=>
=>
=>
=>
This is the required quadratic equation.
Check the picture below.
make sure your calculator is in Degree mode.