A. 2x^2 - 3x + 10 = 2x + 21
2x^2 - 3x - 2x + 10 - 21 = 0
2x^2 - 5x - 11 = 0 (quadratic equation)
2x^2 - 6x - 7 = 2x^2
2x^2 - 2x^2 - 6x - 7 = 0
-6x - 7 = 0 (not a quadratic equation)
5x^2 + 2x - 4 = 2x^2
5x^2 - 2x^2 + 2x - 4 = 0
3x^2 + 2x - 4 (quadratic equation)
5x^3 - 3x + 10 = 2x^2
5x^3 - 2x^2 - 3x + 10 = 0 (not a quadratic equation)
Therefore, options a and c can be solved using the quadratic formula.
Answer:
False
Explanation:
No, a triangle cannot be constructed with sides of 2 in., 3 in., and 6 in.
For three line segments to be able to form any triangle you must be able to take any two sides, add their length and this sum be greater than the remaining side.
we know that
Quotient is the number resulting from the division of one quantity by another
Let
x--------> the first quantity
y------> the second quantity
q------> the quotient
So
-------> equation 1
in this problem
Substitute the values in the equation 1
Simplify
therefore
<u>the answer is</u>
The quotient is equal to 
