Answer:
A: This polynomial has a degree of 2 , so the equation 12x2+5x−2=0 has two or fewer roots.
B: The quadratic equation 12x2+5x−2=0 has two real solutions, x=−2/3 or x=1/4 , and therefore has two real roots.
Step-by-step explanation:
f(x) = 12x^2 + 5x - 2.
Since this is a quadratic equation, or a polynomial of second degree, one can easily conclude that this equation will have at most 2 roots. At most 2 roots mean that the function can have either 2 roots at maximum or less than 2 roots. Therefore, in the A category, 2nd option is the correct answer (This polynomial has a degree of 2 , so the equation 12x^2 + 5x − 2 = 0 has two or fewer roots).
To find the roots of f(x), set f(x) = 0. Therefore:
12x^2 + 5x - 2 = 0. Solving the question using the mid term breaking method shows that 12*2=24. The factors of 24 whose difference is 5 are 8 and 3. Therefore:
12x^2 + 8x - 3x - 2 = 0.
4x(3x + 2) -1(3x+2) = 0.
(4x-1)(3x+2) = 0.
4x-1 = 0 or 3x+2 = 0.
x = 1/4 or x = -2/3.
It can be seen that f(x) has two distinct real roots. Therefore, in the B category, 1st Option is the correct answer (The quadratic equation 12x2+5x−2=0 has two real solutions, x=−2/3 or x=1/4 , and therefore has two real roots)!!!
Answer:
Hi
Step-by-step explanation:
Is there like a multiple choice so I could answer it:)
Equation of the line:
y=mx+b
Ordinate at the origin (where the line cut the y-axis): b=1
B=(-5,5)=(xb,yb)→xb=-5, yb=5
C=(0,1)=(xc,yc)→xc=0, yc=1
m=(yc-yb)/(xc-xb)
m=(1-5)/(0-(-5))
m=(-4)/(0+5)
m=(-4)/(5)
m=-(4/5)
y=mx+b
y=-(4/5)x+1
Multiplying the equation by 5:
5[y=-(4/5)x+1]
5y=-4x+5
Adding 4x both sides of the equation:
5y+4x=-4x+5+4x
4x+5y=5
Answer: The equation for BC is: Third option 4x+5y=5
<span>Rational
expressions are defined as the quotients of two integers in which the
denominator does not equal to zero. Thus, the numerator and denominator would
be polynomials, which is a form of a system analogous to the integers under the
operation of addition, subtraction and multiplications. Therefore, the key
numbers of rational expressions are its variables in the numerator and/ or
denominator.</span>
