There are 38 five point questions and 30 two points questions.
Step-by-step explanation:
Given,
Worth of points in test = 250
Number of questions = 68
Let,
x represent the number of 5 points questions
y represent the number of 2 points question
According to given statement;
x+y=68 Eqn 1
5x+2y=250 Eqn 2
Multiplying Eqn 1 by 2
Subtracting Eqn 3 from Eqn 2
Dividing both sides by 3
Putting x=38 in Eqn 1
There are 38 five point questions and 30 two points questions.
Keywords: linear equation, elimination method
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Yes 373 is in fact a prime number
Answer:
The quadratic x2 − 5x + 6 factors as (x − 2)(x − 3). Hence the equation x2 − 5x + 6 = 0
has solutions x = 2 and x = 3.
Similarly we can factor the cubic x3 − 6x2 + 11x − 6 as (x − 1)(x − 2)(x − 3), which enables us to show that the solutions of x3 − 6x2 + 11x − 6 = 0 are x = 1, x = 2 or x = 3. In this module we will see how to arrive at this factorisation.
Polynomials in many respects behave like whole numbers or the integers. We can add, subtract and multiply two or more polynomials together to obtain another polynomial. Just as we can divide one whole number by another, producing a quotient and remainder, we can divide one polynomial by another and obtain a quotient and remainder, which are also polynomials.
A quadratic equation of the form ax2 + bx + c has either 0, 1 or 2 solutions, depending on whether the discriminant is negative, zero or positive. The number of solutions of the this equation assisted us in drawing the graph of the quadratic function y = ax2 + bx + c. Similarly, information about the roots of a polynomial equation enables us to give a rough sketch of the corresponding polynomial function.
As well as being intrinsically interesting objects, polynomials have important applications in the real world. One such application to error-correcting codes is discussed in the Appendix to this module.
Answer:
just one.. so A
Step-by-step explanation:
