Answer:
The sine of the acute angle is
Step-by-step explanation:
using a graphing tool
see the attached figure
Let
A-----> the acute angle formed by the given line and the positive x-axis
In the right triangle XYZ
we know that
we have
Applying the Pythagoras Theorem find the value of xz
substitute
Induced emf = -NA (ΔФ/Δt)
N = 1 (assuming its 1 loop), ΔФ = 0.2-0 = 0.2 T, Δt = 0.3 sec, A = 0.12 m^2
Therefore,
Induced emf = -1*0.12*(0.2/0.3) = -0.08 V
Answer:
D is correct option
Step-by-step explanation:
The correct option is D.
The standard quadratic equation is ax²+bx+c=0
Where a and b are coefficients and c is constant.
It means that constant are on the L.H.S and there is 0 on the right hand side.
Therefore to make it a quadratic equation first of all you have to add 11 at both sides so that the R.H.S becomes 0.
The given equation is:
2x2-x+ 2 = -11
If we add 11 on both sides the equation will be:
2x2-x+ 2 +11= -11+11
2x^2-x+13=0
Thus the correct option is D
You can further solve it by applying quadratic formula....