Which rules of exponents will be used to evaluate this expression? Check all that apply. StartFraction left-brace (negative 8) S
1 answer:
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Answer: The correct options are (1) (5,10), (2) (3,-3), (3) x = -1, (4) , (5) 21s and (6) 0, -1, and 5.
Explanation:
Te standard form of the parabola is,
.....(1)
Where, (h,k) is the vertex of the parabola.
(1)
The given equation is,
Comparing this equation with equation (1),we get,
and
Therefore, the vertex of the graph is (5,10) and the fourth option is correct.
(2)
The given equation is,
To make perfect square add , i.e.,
. Since there is factor 3 outside the parentheses, so subtract three times of 9.
Comparing this equation with equation (1),we get,
and
Therefore, the vertex of the graph is (3,-3) and the fourth option is correct.
(3)
The given equation is
To make perfect square add , i.e.,
. Since there is factor 4 outside the parentheses, so subtract three times of 1.
Comparing this equation with equation (1),we get,
and
The vertex of the equation is (-1,3) so the axis is x=-1 and the correct option is 2.
(4)
The given equation is,
To make perfect square add , i.e.,
.
Therefore, the correct option is 4.
(5)
The given equation is,
It can be written as,
It is a downward parabola. so the maximum height of the function is on its vertex.
The x coordinate of the vertex is,
Therefore, after 21 seconds the projectile reach its maximum height and the correct option is first.
(6)
The given equation is,
Use factoring method to find the factors of the equation.
Equate each factor equal to 0.
Therefore, the zeros of the given equation is 0, -1, 5 and the correct option is 2.
The smallest such number is 1055.
We want to find such that
The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.
Taking everything together, we end up with the system
Now the moduli are coprime and we can apply the CRT.
We start with
Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.
Taken modulo 2, we end up with
which means the first term is fine and doesn't require adjustment.
Taken modulo 3, we have
We want a remainder of 2, so we just need to multiply the second term by 2.
Taken modulo 5, we have
We want a remainder of 0, so we can just multiply this term by 0.
Taken modulo 7, we have
We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since , the inverse of 2 is 4.
So, we have to adjust to
and from the CRT we find
so that the general solution for all integers
.
We want a 4 digit solution, so we want
which gives .